Natural Applications of Colloids:

Blue Colour of Sky:

When the light is incident on particles whose size is smaller than the wavelength of light, it is scattered. The blue colour of the sky is due to the scattering of light by small particles (dust particles along with water) of the atmosphere. According to Rayleigh the intensity of scattered light is inversely proportional to the fourth power of wavelength. As the wavelength of blue colour is smallest and that of red light is longest, the blue light is scattered most and the red light is scattered the least. The scattered blue light reaching the eye gives the appearance of a blue sky.

When the aeroplane is flying high where there are no dust particles or water vapour then no scattering of any colour takes place and the sky looks black.

Red and Orange Colour of Sky in The Sunrise and Sunset:

When the sun is low on the horizon,  sunlight has to cover more distance through the atmosphere at sunset and sunrise than during the day, when the sun is higher in the sky. More distance through atmosphere means more molecules to scatter the violet and blue light away from our eyes. If the path is long enough, all of the blue and violet light scatters out of our line of sight. But the other colours continue on their way to our eyes. This is why sunrise and sunsets are often yellow, orange, and red.

Blue Colour of Sea:

The colloidal impurities in seawater due to their smaller size less than the wavelength of light scatter blue light. The Tyndall effect of scattering of light by colloids is responsible for the blue colour of the sea.

Fog Mist and Rain:

Fog, mist and rain are all colloidal in nature. In winters, at night, the moisture in the air condenses on the surface of dust particles, forming tiny droplets. These droplets, being colloidal in nature, float in the air and forms mist or fog.

Clouds are aerosols consisting of small droplets of water suspended in the air (aerosols). Clouds are the colloidal solution. On account of condensation in the upper atmosphere, the colloidal particles of water become bigger and bigger till they come down in the form of rain. They carry an electrical charge. The condensation occurs when the dust particles are cooled below its dew point. Sometimes rainfall occurs when oppositely charged clouds meet.

In the artificial rain, the clouds are sprayed by oppositely charged colloidal dust or sand particles or precipitates of silver iodide. This spraying neutralizes the charge on cloud resulting in coagulation of the water droplets which come down in the form of rain.

Industrial Applications of Colloids:

Thickening Agents:

Toothpaste, lotions, lubricants, coatings are the substances where the viscosity (degree of flow-ness) is very important. The substances added to them to change and maintain the viscosity are colloidal in nature. These colloid particles also provide stabilization of the colloidal solution and prevent the phase separation. They also act as fillers.

Example: various natural gums, microcrystalline cellulose, carboxymethyl cellulose, and fumed silica.

Paints:

Paints have been used since ancient times for both protective and decorative purposes. They consist basically of pigment particles dispersed in a liquid. The liquid capable of forming a stable solid film as the paint “dries”.on drying of the paint. On exposure to air, the pigments polymerize into the impervious film.

Inks:

The most critical properties of inks relate to their drying and surface properties. Inks must be able to flow properly and attach to the surface without penetrating it. It should dry very fast. Many inks consist of organic dyes dissolved in a water-based solvent and are not colloidal at all. The ink used in printing newspapers employs colloidal carbon black dispersed in an oil as the dispersion medium.

The inks employed in ball-point pens are gels, made in such a way that the ink will only flow over the ball and onto the paper when the shearing action of the ball (which rotates as it moves across the paper) “breaks” the gel into a liquid; the resulting liquid coats the ball and is transferred to the paper.

In conventional printing, the pigment particles remain on the paper surface, while the liquid gradually evaporates.

Rubber:

Latex obtained from rubber trees is an emulsion consisting of negatively charged rubber particles in water. Rubber is obtained by the coagulation of latex. This coagulated mass is later subjected to vulcanization and is solid as rubber with high abrasive strength. Vulcanized rubber is used in making tyres for vehicles.

Rubber plated articles can be prepared directly from latex by electrically depositing the negat6ively charged rubber particles over the article (made anode) which is to be rubber plated.

Leather Tanning Industry:

Raw skin hides of animals contain positively charged colloidal particles. These particles are coagulated by negatively charged tannin materials. After the tanning process, the leather becomes harder. Tanning material used are tannin and compounds of aluminium and chromium.

Cleansing Action of Soaps:

Soap solutions are colloidal in nature. They remove the dirt and oil particles either by adsorption or by emulsifying the greasy matter sticking to cloth.

Disinfectants:

Dettol and Lysol form an oil in water type colloidal solution which is used as the disinfecting agent.

Metallurgy:

The colloidal mixture of oil in water is used in the froth flotation process to separate sulphide ore particles.

Making of Photographic Plates:

Photographic plates and films are produced by coating an emulsion of the light-sensitive material like silver bromide in gelatin over class plates or celluloid films.

Sewage Precipitation:

Dirty and muddy water from gutters and drainages is called sewage is in colloidal form (colloidal solution).

Sewage water containing colloidal particles of mud, rubbish etc. is collected in a tank fitted with electrodes.

On applying an electric field, colloidal particles are attracted towards oppositely charged electrodes. As their charge gets neutralised, they settle as a precipitate. The precipitated or coagulated matter called sludge is used as manure while clear water is used for irrigation.

Smoke Precipitation:

Smoke is a colloidal solution of negatively charged carbon particles in the air (aerosol)

These carbon particles may condense water vapour on them and thus cities may have a thick cover of smog (smoke + fog). This smog causes air pollution.

Cottrel’s precipitator is a widely used smoke precipitator. Smoke is passed between metal electrodes at high voltage (about 50,000 V) The charged particles are neutralized at the oppositely charged electrode and get deposited there. The gases free from carbon particles are passed to a chimney or for further purification.

Other Applications of Colloids:

Synthetic plastics, rubber, graphite, lubricants, cement, etc. are colloidal solutions. Asphalt emulsion is used for road construction. The principles of colloids and interface science are used for the successful formulation and manufacture of photographic products.

Applications of Colloids in Food:

Food:

Most of the foods we eat are largely colloidal in nature. The function of food colloids generally has less to do with nutritional value than appearance, texture, and “mouthfeel”.  Mouthfeel is the ability to “melt” (transform from gel to liquid emulsion) on contact with the warmth of the mouth.

Dairy Products:

milk (oil in water), butter (water in oil), halva, icecreams, are in colloidal form.

Fruit Juices:

Fruit juices are colloidal solutions

Eggs:

The clear, viscous “white” of the egg can be transformed into a white, opaque semi-solid by brief heating, or rendered into more intricate forms by poaching, frying, scrambling, or baking.

The raw egg white is a colloidal sol of long-chain protein molecules, all curled up into compact folded forms due to hydrogen bonding between different parts of the same molecule. On heating, the hydrogen bonds are broken, and proteins unfold. The opened chains tangle and bind with each other, transforming the sol into a cross-linked hydrogel, and changes its appearance to opaque white.

Applications of Colloids in Agriculture:

Soil:

The four major components of soils are mineral sediments, organic matter, water, and air.  A fertile soil is colloidal in nature in which humus acts as a protective colloid.

Most soil colloids are negatively charged, and therefore attract cations such as Ca²⁺, Mg²⁺, and K⁺ into the outer parts of their double layers. As these ions are loosely bound, the plant roots can absorb these essential nutrients. Similarly, these ions are released into the soil again when the plant dies.

Formation of Delta:

River water is a colloidal solution of clay and seawater which mainly carry a negative charge. Seawater contains different electrolytes, mainly positive ions Na⁺, Mg²⁺, Ca²⁺. When the river meets the sea, the electrolytes in seawater bring about coagulation of clay particles in the river water. Thus the clay particles aggregate and settle down in course of water. Which results in the formation of a delta in due course.

Applications of Colloids in Medicine:

Colloidal medicines can act on a large surface area, hence they are more easily assimilated by the body. Therefore, they are more effective

Argyrol, a silver metal sol used as an eye lotion, colloidal antimony used for curing Kalaazar, colloidal gold used for inter-muscular injections.

Human Physiology:

Blood is a colloidal solution of albuminoid substances. Bleeding from a fresh cut can be stopped by applying a concentrated solution of ferric chloride or potash alum (this is known as the styptic action of alum or ferric chloride). In this case the cpagulation of blood takes place and a clot is formed which prevents further bleeding.

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In thisarticle, we shall study bout associated colloids :micelles).

Multimolecular Colloids:

Multimolecular colloids are those systems in which the dispersed phase particles are aggregates of many atoms or molecules. The particles in this colloidal solutions are held together by van der Wall’s forces. e.g. gold sol particles are an aggregation of many gold atoms. Other examples are silver sol and sulphur sol.

Macromolecular Colloids:

Macromolecular colloids are those systems in which the dispersed phase particles are a single macromolecule. They are lyophilic in character. e.g. sol of starch in water, Aqueous (Water) solution of proteins, enzymes.

Associated Colloids or Micelles

Colloids which behave as normal electrolytes at low concentrations, but exhibit colloidal properties at higher concentrations due to the formation of aggregated particles called associated colloids. The aggregated particles thus formed are called micelles.

The associated colloids are usually formed by surfactants (surface active agents) like soaps and synthetic detergents. The molecules of soaps and detergents are smaller than the colloidal particles. When dissolved in water soap and detergent molecules act as an electrolyte but if their concentration is increased then their molecules aggregate to form colloidal size particles called micelles. The formation of micelles takes place only above a certain concentration, this concentration is known as critical micellization concentration (CMC).

The Process of Formation of Micelles:

The formation of a micelle can be understood by taking the example of a soap solution. In general, soap can be represented as RCOONa, where R represents a long chain alkyl group. When dissolved in water, soap ionizes to give RCOO– and Na⁺ ions. The most commonly used washing soap is sodium stearate, C₁₇H₃₅COONa.

Clothes become dirty due to the deposition of dust and oily or greasy substances. Water is not capable of wetting oily or greasy substances. However, the hydrocarbon residue R of the soap anion (RCOO^–) can wet the oily or greasy substances. Soaps and detergents have two parts, the hydrophobic part and the hydrophilic part.

Hydrophobic or water repelling non-polar part (usually a long hydrocarbon chain) is soluble in oil and greases but insoluble in water. Hydrophilic or water attracting polar parts such as carboxylic group or sulphonate or sulphate is soluble in water and insoluble in oil and greases.

Molecules of soap and detergent form micelles in water. The hydrophobic part of soap dissolves in oil and grease while hydrophilic part of soap remains as free in soap solution. When the cloth is rubbed with hand or stirred mechanically, the big molecules of oil and soap break into small emulsified oil droplets.  These oil droplets repel each other due to the presence of anions of the hydrophilic part and do not precipitate. Thus they remain suspended in the soap solution without getting back on the cloth. These suspended oil and grease particles are then washed away by a stream of water.

In micelle formation, the long hydrocarbon chain which is insoluble in water is directed towards the centre while the soluble polar head is on the surface in contact with water.

The cleaning action of detergents such as sodium lauryl sulphate, CH₃ (CH₂)₁₁SO₄Na⁺ or sulphonates of long-chain hydrocarbons is similar to that of soaps. In case of detergents, the polar groups are sulphate (–SO₄) or sulphonate (-SO₃) groups.

Kraft’s Temperature:

The formation of micelles takes place only above a particular temperature is called the kraft’s temperature.

More Examples of Micelle System:

Sodium lauryl sulphate:

Sodium oleate:

Cetyltrimethyl ammonium bromide:

It forms micelle with a cationic terminal.

p-Dodecyl benzene sulphonate:

Surfactants:

Surfactants are compounds that lower the surface tension. They are preferentially adsorbed at the interfaces between two liquids, between a gas and a liquid, or between a liquid and a solid. Surfactants may act as detergents, wetting agents, emulsifiers, foaming agents, and dispersants.

Cationic Surfactants:

The cationic surfactants are quaternary ammonium compounds with aryl or alkyl substituent groups, one of which is often a long hydrophobic carbon chain. Cationic surfactants are positively charged, and therefore are not as effective as detergents in cleansing systems. They can ideally be used as fabric softener.

Cetyl pyridinium chloride:

Cetyltrimethyl ammonium chloride:

Octadecyl ammonium chloride

Anionic Surfactants:

Anionic surfactant gives anion. Anionic surfactants are positively charged and are widely used.  Anionic surfactants possess a negative charge on their hydrophilic end. Generally, they make a lot of foam when agitated. They are free-flowing powdery when dry, not sticky like other surfactants.

Sodium Palmitate:

Sodium oleate:

Salts of sulphonic acid

Non-ionogenic or Non-ionic Surfactants:

Nonionic surfactants have no charge on their hydrophilic end, hence they are used as superior oily soil emulsifiers. They do not ionise or dissociate in an aqueous medium. Because of their lower foam profile and strong emulsifying potential, these surfactants are the preferred choice when formulating extraction cleaners and pre-sprays. Nonionics are thick liquids or syrups that are sticky. Nonionic surfactants include: ethoxylates, alkoxylates and cocamide

Characteristics of Micelles

• The formation of micelles takes place only above a certain concentration, this concentration is known as critical micellization concentration (CMC). As the concentration of a surfactant increases, adsorption takes place at the surface until it is fully overlaid, which corresponds to the minimum value of the surface tension.
• An increase in temperature usually increases CMC.
• Greater the chain length of the hydrocarbon chain smaller is the CMC.
• An increase in the hydrophobic part of the surfactant increases CMC.
• The addition of simple electrolyte in ionic micelles decreases their CMC.
• The formation of micelles takes place only above a particular temperature is called the Kraft’s temperature. Below Kraft’s temperature the solubility of surfactant not enough to form micelles.
• Kraft’s temperature increases with the increase in the number of carbon atoms.

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In this article, we shall study types of colloidal solutions (systems) on the basis of states of the dispersed phase and dispersion medium, the interaction between the dispersed phase and dispersion medium, and on the number of atoms and molecules in a colloidal particle.

Types of Colloidal Solutions on the Basis of States of Dispersed Phase and Dispersion Medium:

A colloidal system is made up of a dispersed phase and a dispersion medium. Because either the dispersed phase or the dispersion medium can be a gas, liquid or solid. There are eight types of the colloidal system possible since gases are miscible, the gas colloidal system is not possible. Gas-gas systems always form true solutions.

Classification of Sols On The Basis of  Interaction Between the Dispersed Phase and the Dispersion Medium:

A colloidal solution in which the dispersed phase is in solid-state and the dispersion medium is liquid is called a sol. e.g. Gum solution, the starch in water, Au, Ag, etc. in water, blood, etc.

Lyophilic Sols or Reversible Sols (Emulsoid):

The sols in which there is a strong affinity between the dispersed phase and dispersion medium are called as lyophilic sols. e.g. glue, gelatin, starch, proteins.

Characteristics of Lyophilic Sols:

• Lyophilic sols are readily formed by mixing together the substance forming disperse phase and solvent forming dispersion medium and heating the mixture if necessary.
• They are stable.
• There is a strong affinity between the dispersed phase and dispersion medium.
• The colloidal particles forming lyophilic sols are large single molecules or polymers like starch, proteins etc. of high molecular weight.
• If lyophilic sol is heated or dried we get solid but we get same sol if liquid (solvent or dispersion medium) is added to the solid. Thus lyophilic sols are reversible. After coagulation, they can again be converted into colloidal form.
• Lyophilic sols have lower surface tension than the dispersion medium.
• Lyophilic sols have a higher viscosity than dispersion medium.
• Stability of Lyophilic sols is due to high solvation due to the high affinity of particles towards dispersing medium.
• Lyophilic sols are stable and require a large quantity of electrolyte for coagulation. Thus they can not be coagulated easily.
• The particles cannot be detected easily under ultramicroscope.
• Lyophilic sols show weak Tyndall effect.

Stability of Lyophilic Sols:

In lyophilic sol, a thin film of the dispersion medium is formed around the dispersed phase colloidal particles due to the strong affinity between the dispersed phase and dispersion medium. The formation of this film around dispersed phase colloidal particles is called solvation. The stability of lyophilic sol is due to solvation.

Similarly, all the particles carry an electrical charge of the same nature, which results in mutual repulsion between the dispersed phase colloidal particles which also adds to the stability of lyophilic sol. But the charge on particles is very less or almost negligible.

Thus the stability of Lyophilic sols is due to solvation and charge on colloidal particles.

Lyophobic Sols or Irreversible Sols:

The sols in which there is no affinity between the dispersed phase and dispersion medium are called as lyophobic sols. e.g. sols of metals like Ag, Au, non-metals like sulphur, hydroxides like Al(OH)₃, Fe(OH)₃, sulphides like As₂S₃.

Characteristics of Lyophobic Sols:

• Lyophobic sols cannot be readily formed by mixing together the substance forming disperse phase and solvent forming dispersion medium.  Special methods like dispersion method or condensation method should be employed for making lyophobic sols.
• They are less stable.
• There is no or very little affinity between the dispersed phase and dispersion medium.
• The colloidal particles forming lyophobic sols are aggregates of a large number of atoms or molecules.
• If lyophilic sol is evaporated we get solid but we can not get the same sol if liquid (solvent or dispersion medium) is added to the solid. Thus lyophobic sols are irreversible. After coagulation, they cannot be converted into colloidal form again.
• Lyophobic sols have the same surface tension as the dispersion medium.
• Lyophobic sols have the nearly same viscosity as the dispersion medium.
• The stability of lyophobic sol is due to the charge on colloidal particles.
• Lyophobic sols are unstable and require a very small quantity of electrolyte for coagulation. Thus can be coagulated easily.
• The particles can be detected easily under an ultramicroscope.
• Lyophobic sols show a strong Tyndall effect.

Stability of Lyophobic Sols:

In lyophobic sols, all colloidal particles of the dispersed phase are either positively charged or negatively charged.

Colloidal particles remain suspended in the dispersion medium, without coagulation due to the repulsion between the particle having same nature of the charge,

Thus the stability of lyophobic sol is due to charge on colloidal particles.

Notes:

• If the dispersion medium is water then lyophilic and lyophobic sols are called hydrophilic and hydrophobic sols respectively.
• The colloidal solutions in alcohol and benzene are known as alcosols and benzosols respectively.
• The colloidal solutions in water are known as aquasols or hydrosols.

Classification of Colloidal Solutions on the Basis of the Number of Molecules or Atoms in the Colloidal Particle:

Multimolecular Colloids:

Multimolecular colloids are those systems in which the dispersed phase particles are aggregates of many atoms or molecules. The particles in these colloidal solutions are held together by van der Wall’s forces. e.g. gold sol particles are an aggregation of many gold atoms. other examples are silver sol and sulphur sol.

Macromolecular Colloids:

Macromolecular colloids are those systems in which the dispersed phase particles are a single macromolecule. They are lyophilic in character. e.g. sol of starch in water, Aqueous (Water) solution of proteins, enzymes.

Associated Colloids:

Colloids which behave as normal electrolytes at low concentrations, but exhibit colloidal properties at higher concentrations due to the formation of aggregated particles called associated colloids. The aggregated particles thus formed are called micelles.

Science > Chemistry > Colloids >Types of Colloidal Solutions

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In this article, we shall study two types of solutions first, those obeying Raoult’s law called ideal solutions and those not obeying Raoult’s law called non-ideal solutions.

Ideal Solutions:

The solutions which obey Raoult’s law over the entire range of concentration are known as ideal solutions.

The ideal solutions have two important properties. The enthalpy of mixing of the pure components to form the solution is zero and the volume of mixing is also zero, i.e. ΔmixH = 0 and ΔmixV = 0. It means that no heat is absorbed or evolved when the components are mixed. Also, the volume of solution would be equal to the sum of volumes of the two components.

At the molecular level, ideal behaviour of the solutions can be explained by considering two components A and B. In pure components, the intermolecular attractive interactions will be of types A-A and B-B, whereas in the binary solutions in addition to these two interactions, A-B type of interactions will also be present.

If the intermolecular attractive forces between the A-A and B-B are nearly equal to those between A-B, this leads to the formation of an ideal solution.

A perfectly ideal solution is rare but some solutions are nearly ideal in behaviour. The solution of n-hexane and n-heptane, bromoethane and chloroethane, benzene and toluene, chlorobenzene and bromobenzene etc. fall into this category. Most of the dilute solutions behave as ideal solutions.

Notes:

The process of separation of one liquid from another liquid (binary mixture) having different boiling points by distillation is called fractional distillation. The separation is possible when the vapour phase has a different composition from that boiling liquid mixture. Thus the components of an ideal solution can be separated by fractional distillation.

Examples of ideal solutions:

• All dilute solutions
• benzene + toluene
• n-hexane + n-heptane
• chlorobenzene + bromobenzene
• ethyl bromide + ethyl iodide
• n-butyl chloride + n-butyl bromide

Non-ideal Solutions:

When a solution does not obey Raoult’s law over the entire range of concentration, then it is called a non-ideal solution.

The vapour pressure of such a solution is either higher or lower than that predicted by Raoult’s law. If it is higher, the solution exhibits a positive deviation and if it is lower, it exhibits a negative deviation from Raoult’s law.

Positive Deviation:

The cause for these deviations lies in the nature of interactions at the molecular level. In case of positive deviation from Raoult’s law, A-B interactions are weaker than those between A-A or B-B, i.e.

In this case, the intermolecular attractive forces between the solute-solvent molecules are weaker than those between the solute-solute and solvent-solvent molecules. This means that in such solutions, molecules of A (or B) will find it easier to escape than in the pure state. This will increase the vapour pressure and result in positive deviation.

Mixtures of ethanol and acetone behave in this manner. In pure ethanol, molecules are hydrogen-bonded. On adding acetone, its molecules get in between the host molecules and break some of the hydrogen bonds between them. Due to the weakening of interactions, the solution shows a positive deviation from Raoult’s law.

In a solution formed by adding carbon disulphide to acetone, the dipolar interactions between solute-solvent molecules are weaker than the respective interactions among the solute-solute and solvent-solvent molecules. This solution also shows positive deviation.

For solution showing positive deviation from Rault’s law ΔmixH > 0 and ΔmixV > 0,

Examples of solutions showing positive deviation from Raoult’s law

• acetone + ethanol
• acetone + CS2
• water + methanol
• water + ethanol
• CCl4 + toluene
• CCl4 + CHCl3
• acetone + benzene
• CCl4+ CH3OH
• cyclohexane + methanol

Negative Deviation:

In the case of negative deviations from Raoult’s law, the intermolecular attractive forces between A-A and B-B are weaker than those between A-B and leads to decrease in vapour pressure resulting in negative deviations.

An example of this type is a mixture of phenol and aniline. In this case, the intermolecular hydrogen bonding between phenolic proton and lone pair on the nitrogen atom of aniline is stronger than the respective intermolecular hydrogen bonding between similar molecules. Similarly, a mixture of chloroform and acetone forms a solution with a negative deviation from Raoult’s law. This is because chloroform molecule is able to form a hydrogen bond with acetone molecule as shown.

This decreases the escaping tendency of molecules for each component and consequently the vapour pressure decreases resulting in a negative deviation from Raoult’s law. This decreases the escaping tendency of molecules for each component and consequently the vapour pressure decreases resulting in a negative deviation from Raoult’s law.

For solution showing negative deviation from Rault’s law ΔmixH < 0 and ΔmixV < 0,

Examples of solutions showing negative deviation from Raoult’s law

• acetone + aniline
• acetone + chloroform
• methanol + acetic acid
• water + nitric acid
• chloroform + diethyl ether
• water + HCl
• acetic acid + pyridine
• chloroform + benzene

Azeotropes:

Azeotropes are binary mixtures having the same composition in the liquid and vapour phase and boil at a constant temperature. Some liquids on mixing, form azeotropes. In such cases, it is not possible to separate the components by fractional distillation. There are two types of azeotropes called minimum boiling azeotrope and maximum boiling azeotrope.

The solutions which show a large positive deviation from Raoult’s law form minimum boiling azeotrope at a specific composition. For example, ethanol-water mixture (obtained by fermentation of sugars) on fractional distillation gives a solution containing approximately 95% by volume of ethanol. Once this composition, known as azeotrope composition, has been achieved, the liquid and vapour have the same composition, and no further separation occurs.

The solutions that show large negative deviation from Raoult’s law form maximum boiling azeotrope at a specific composition. Nitric acid and water is an example of this class of azeotrope. This azeotrope has the approximate composition, 68% nitric acid and 32% water by mass, with a boiling point of 393.5 K.

Related Topics

Solutions and Their Colligative Properties

• Solutions and Their Types
• Solutions of Solids and Liquids
• Concentration of Solution
• Numerical Problems on Percentage by Mass and Volume
• Numerical Problems on Mole Fraction
• Numerical Problems on Molarity
• Numerical Problems on Molality
• Short Cuts For Above Numerical Problems
• Solutions of Gases in Liquid
• Lowering of Vapour Pressure
• Numerical Problems on Lowering of Vapour Pressure
• Elevation in Boiling Point and Depression in Freezing Point
• Osmosis and Osmotic Pressure

For More Topics of Chemistry Click Here

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Science > Chemistry > Solutions and Their Colligative Properties > Short-cut Methods For Calculating Concentration of Solutions

In this article, we shall study short-cut methods to calculate molality, molarity, etc.

These methods can only be used in competitive exams only.

Direct Formulae to Calculate Molality and Molarity:

Where M = molarity in mol L^-1 or M

m = molality in mol kg^-1 or m

a = % by mass of solute

d = density of solution in g/mL or g cm^-3.

M_B = Molecular mass of solute in grams

M_A = Molecular mass of solvent in grams

Note: When using these formulae, take care that the quantities are in prescribed units

Molecular masses of certain substances in grams:

Water H₂O (18), Benzene C₆H₆ (78), Sodium hydroxide NaOH (40), Hydrogen chloride HCl (36.5), Sulphuric acid H₂SO₄ (98), potassium hydroxide KOH (56), Acetic acid (60), Sodium carbonate Na₂CO₃ (116),

Numerical Problems to Calculate Molality and Molarity:

Example – 01:

The density of a solution containing 13 % by mass of sulphuric acid is 1.09 g/mL. Calculate molarity and normality of the solution

Given: a = 13, d = 1.09 g/mL

To Find: Molarity (M) =? and Normality (N) =?

Solution:

n = Molecular mass/equivalent mass = 98 g/49 g = 2

Normality = molarity x n = 1.446 x 2 = 2.892 N

Example – 02:

The density of 2.03 M solution of acetic acid (molecular mass = 60) in water is 1.017 g/mL. Calculate molality of solution

Given: M = 2.03, M_B = 60 g mol^-1, d = 1.017 g/mL

To Find: Molality (m) = ?

Solution:

molality = m = 1/0.4410 = 2.268 molal

Example – 03:

The density of 10.0% by mass of KCl solution in water is 1.06 g/mL. Calculate the molality, molarity and mole fraction of KCl.

Given: a = 10, d = 1.06 g/mL

To Find: Molarity (M) =?, molality (m) =?, mole fraction (X_B) =?

Solution:

Ans: Molarity 1.42 M, Molality = 1.491 m, Mole fraction = 0.0261

Example – 04:

0.8 M solution of H2SO4 has a density of 1.06 g/cm³. calculate molality and mole fraction

Given: M = 0.8 M, d = 1.06 g/cm³.

To Find: Molality (m) =?, mole fraction (X_B) =?

Solution:

molality = m = 1/1.227 = 0.814 molal

0.814 x 18 x (1 – X_B) = 1000 X_B

14.652 – 14.652 X_B = 1000 X_B

1014.652 X_B = 14.652

X_B = 14.652/1014.652  = 0.014

Example – 05:

A 6.90 M solution of KOH in water contains 30% by mass of KOH. Calculate the density of solution.

Given: M = 6.90 M, a = 30

To Find: density of solution = d = ?

Solution:

Ans: Density of solution = 1.288 g/mL

Example – 06:

An aqueous solution of NaOH is marked 10% (w/w). The density of the solution is 1.070 g cm^-3. Calculate molality, molarity and mole fraction of NaOH in water. Given Na = 23, H =1 , O = 16

Given: a = 10, d =  1.070 g cm^-3,

To Find: mole fraction =? molarity = ? and molality =?

Solution:

Example – 07:

Calculate the mole fraction of solute in its 2 molal aqueous solution.

Given: molality = 2 molal

To Find: Mole fraction =?

Solution:

Related Topics

Solutions and Their Colligative Properties

• Solutions and Their Types
• Solutions of Solids and Liquids
• Concentration of Solution
• Numerical Problems on Percentage by Mass and Volume
• Numerical Problems on Mole Fraction
• Numerical Problems on Molarity
• Numerical Problems on Molality
• Solutions of Gases in Liquid
• Ideal and Non-ideal Solutions
• Lowering of Vapour Pressure
• Numerical Problems on Lowering of Vapour Pressure
• Elevation in Boiling Point and Depression in Freezing Point
• Osmosis and Osmotic Pressure

For More Topics of Chemistry Click Here

The post Short-cut Methods For Calculating Concentration of Solutions appeared first on The Fact Factor.

In this article, we shall study numerical problems to calculate molality of a solution.

Example – 01:

7.45 g of potassium chloride (KCl) was dissolved in 100 g of water. Calculate the molality of the solution.

Given: mass of solute (KCl) = 7.45 g, mass of solvent (water) = 100 g = 0.1 kg

To Find: Molarity of solution =?

Solution:

Molecular mass of KCl = 39 g x 1 + 35.5 g x 1 = 74.5 g mol^-1

Number of moles of solute (KCl) = given mass/ molecular mass

Number of moles of solute (KCl) = 7.45 g/ 74.5 g mol^-1 = 0.1 mol

Molality = Number of moles of solute/Mass of solvent in kg

Molality = 0.1 mol /0.1 kg = 1 mol kg^-1

Ans: The molality of solution is 1 mol kg^-1 or 1 m.

Example – 02:

11.11 g of urea (NH₂CONH₂) was dissolved in 100 g of water. Calculate the molarity and molality of the solution. Given N = 14, H = 1, C = 12, O = 16.

Given: mass of solute (urea) = 11.11 g, mass of solvent (water) = 100 g = 0.1 kg

To Find: Molarity of solution =?

Solution:

Molecular mass of urea (NH₂CONH₂) = 14 g x 2 + 1 g x 4 + 12 g x 1 + 16 g x 1

Molecular mass of urea (NH₂CONH₂) = 60 g mol^-1

Number of moles of solute (urea) = given mass/ molecular mass

Number of moles of solute (urea) = 11.11 g/ 60 g mol^-1 = 0.1852 mol

Volume of water = mass of water/ density = 100 g/1 g mL^-1 = 100 mL = 0.1 L

Molarity = Number of moles of solute/Volume of solution in L

Molarity = 0.1852 mol /0.1 L = 1.852 mol L^-1 or 1.852 mol dm^-3

Molality = Number of moles of solute/Mass of solvent in kg

Molality = 0.1852 mol /0.1 kg = 1.852 mol kg^-1

Ans: The molarity of solution is 1.852 mol L^-1 and the molality is 1.852 mol kg^-1

Example – 03:

34.2 g of sugar was dissolved in water to produce 214.2 g of sugar syrup. Calculate molality and mole fraction of sugar in the syrup. Given C = 12, H = 1 and O = 16.

Given: Mass of solute (sugar) = 34.2 g, Mass of solution (sugar syrup) = 214.2 g

To Find: Molality and mole fraction =?

Solution:

Mass of Solution = Mass of solute + mass of solvent

Mass of solvent = mass of solution – mass of solute = 214.2 g – 34.2 g = 180 g = 0.180 kg

Molar mass of sugar (C₁₂H₂₂O₁₁) = 12 g x 12 + 1 g x 22 + 16 g x 11 = 342 g mol^-1

Number of moles of solute (sugar) = n_B = Given mass/ molecular mass = 34.2 g/342 g mol^-1  = 0.1 mol

Molality = Number of moles of solute/Mass of solvent in kg

Molality = 0.1 mol /0.180 kg = 0.5556 mol kg^-1

Molar mass of water (H₂O) = 1 g x 2 + 16 g x 1 = 18 g mol^-1

Number of moles of solvent (water) = n_A = Given mass/ molecular mass = 180 g/18 g mol^-1  = 10 mol

Total number of moles = n_A + n_B = 0.1 + 10 = 10.1 mol

Mole fraction of solute (sugarl) = x_B = n_B/(n_A + n_B) = 0.1/10.1 = 0.0099

Mole fraction of sugar = 0.0099

Ans: Molality of solution = 0.5556 mol kg^-1 and mole fraction of sugar = 0.0099

Example – 04:

10.0 g KCl is dissolved in 1000 g of water. If the density of the solution is 0.997 g cm^-3, calculate a) molarity and b) molality of the solution. Atomic masses K = 39 g mol^-1, Cl = 35.5 g mol^-1.

Given: the mass of solute (KCl) = 10 g, the mass of solvent (water) = 1000 g = 1 kg, density of solution = 0.997 g cm^-3,

To Find: molarity =? molality = ?

Solution:

Molecular mass of KCl = 39 g x 1 + 35.5 g x 1 = 74.5 g mol^-1

Number of moles of solute (KCl) = given mass/ molecular mass

Number of moles of solute (KCl) = 10 g/ 74.5 g mol^-1 = 0.1342 mol

Molality = Number of moles of solute/Mass of solvent in kg

Molality = 0.1342 mol /1 kg = 0.1342 mol kg^-1

Mass of solution = 10 g + 1000 g = 1010 g

Volume of solution = mass of solution/density = 1010/0.997 g cm^-3

Volume of solution = 1013 cm³ = 1013 mL = 1.013 L

Molarity = Number of moles of solute/Volume of solution in L

Molarity = 0.1342 mol /1.013 L = 0.1325 mol L^-1

Ans: The molarity of the solution is 0.1325 mol L^-1 or 0.1325 M, the molality of the solution is 0.1342 mol kg^-1 or 0.1342 m.

Example – 05:

Calculate molarity and molality of the sulphuric acid solution of density 1.198 g cm^-3 containing 27 % by mass of sulphuric acid.

Given: density of the solution = 1.198 g cm^-3, % mass of sulphuric acid = 27%,

To Find: Molarity =? and molality =?

Solution:

Consider 100 g of solution

Mass of H₂SO₄ = 27 g and mass of H₂O = 100 – 27 g = 73 g = 0.073 kg

Molecular mass H₂SO₄ = 1 g x 2 + 32 g x 1 + 16g  x 4 = 98 g mol^-1

Number of moles of H₂SO₄ = n_B = 27 g/ 98 g = 0.2755 mol

Density of solution = 1.198 g cm^-3

Volume of solution = Mass of solution / density = 100 g /1.198 g cm^-3 = 83.47 cm³ = 83.47 mL = 0.08347 L

Molarity of solution = Number of moles of the solute/volume of solution in L = 0.2755/0.08347 = 3.301 M

Molality = Number of moles of solute/mass of sovent in kg

Molality = 0.2755 mol /0.073 kg = 3.774 mol L^-1

Ans: The molarity of solution is 3.374 mol L^-1 or 3.374 M, the molality of solution is 3.774 mol L^-1 or 3.774 m

Example – 06:

Calculate the mole fraction, molality and molarity of HNO₃ in a solution containing 12.2 % HNO₃. Given density of HNO₃ as 1.038 g cm^-3, H = 1, N = 14, O = 16.

Given: density of the solution = 1.038 g cm^-3, % mass of HNO₃ = 12.2 %,

To Find: mole fraction =? molarity =? and molality =?

Solution:

Consider 100 g of solution

Mass of HNO₃ = 12.2 g and mass of H₂O = 100 – 12.2 g = 87.8 g = 0.0878 kg

Molecular mass of water (H₂O) = 1 g x 2 + 16 g x 1 = 18 g mol^-1

Molecular mass HNO₃ = 1 g x 1 + 14 g x 1 + 16g  x 3 = 63 g mol^-1

Number of moles of water = n_A = 87.8 g/ 18 g = 4.8778 mol

Number of moles of HNO₃ = n_B = 12.2 g/ 63 g = 0.1937 mol

Total number of moles = n_A + n_B + n_C = 4.8778 + 0.1937 = 5.0715

Mole fraction of HNO₃ = x_B = n_B/(n_A +n_B) = 0.1937/5.0715 = 0.0382

Density of solution = 1.038 g cm^-3

Volume of solution = Mass of solution / density = 100 g /1.038 g cm^-3 = 96.34 cm³ = 96.34 mL = 0.09634 L

Molarity of solution = Number of moles of the solute/volume of solution in L = 0.1937/0.09634 =2.011 M

Molality = Number of moles of solute/mass of sovent in kg

Molality = 0.1937 mol /0.0878 kg = 2.206 mol kg^-1

Ans: The mole fraction of HNO3 is 0. 0382, the molarity of solution is 2.011 mol L^-1 or 2.011 M, the molality of solution is 2.206 mol kg^-1 or 2.206 m

Example – 07:

Calculate molarity and molality of 6.3 % solution of nitric acid having density 1.04 g cm^-3. Given atomic masses H = 1, N = 14 and O = 16.

Given: density of the solution = 1.04 g cm^-3, % mass of HNO₃ = 6.3 %,

To Find: mole fraction =? molarity =? and molality =?

Solution:

Consider 100 g of solution

Mass of HNO₃ = 6.3 g and mass of H₂O = 100 – 6.3 g = 93.7 g = 0.0937 kg

Molecular mass of water (H₂O) = 1 g x 2 + 16 g x 1 = 18 g mol^-1

Molecular mass HNO₃ = 1 g x 1 + 14 g x 1 + 16g  x 3 = 63 g mol^-1

Number of moles of water = n_A = 93.4 g/ 18 g = 5.189 mol

Number of moles of HNO₃ = n_B = 6.3 g/ 63 g = 0.1 mol

Density of solution = 1.04 g cm^-3

Volume of solution = Mass of solution / density = 100 g /1.04 g cm^-3 = 96.15 cm³ = 96.15 mL = 0.09615 L

Molarity of solution = Number of moles of the solute/volume of solution in L = 0.1/0.09615 =1.040 M

Molality = Number of moles of solute/mass of sovent in kg

Molality = 0.1 mol /0.0937 kg = 1.067 mol kg^-1

Ans: The molarity of solution is 1.040 mol L^-1 or 1.040 M

The molality of solution is 1.067 mol kg^-1 or 1.067 m

Example – 08:

An aqueous solution of NaOH is marked 10% (w/w). The density of the solution is 1.070 g cm^-3. Calculate molarity, molality and mole fraction of NaOH in water. Given Na = 23, H =1 , O = 16

Given: density of the solution = 1.038 g cm^-3, % mass of HNO₃ = 12.2 %,

To Find: mole fraction =? molarity =? and molality =?

Solution:

Consider 100 g of solution

Mass of NaOH = 10 g and mass of H₂O = 100 – 10 g = 90 g = 0.090 kg

Molecular mass of water (H₂O) = 1 g x 2 + 16 g x 1 = 18 g mol^-1

Molecular mass NaOH = 23 g x 1 + 16 g x 1 + 1 g  x 1 = 40 g mol^-1

Number of moles of water = n_A = 90 g/ 18 g = 5 mol

Number of moles of NaOH = n_B = 10 g/ 40 g = 0.25 mol

Total number of moles = n_A + n_B = 5 + 0.25 = 5.25 mol

Mole fraction of NaOH = x_B = n_B/(n_A +n_B) = 0.25/5.25 = 0.0476

Density of solution = 1.070 g cm^-3

Volume of solution = Mass of solution / density = 100 g /1.070 g cm^-3 = 93.46 cm³ = 93.46 mL = 0.09346 L

Molarity of solution = Number of moles of the solute/volume of solution in L = 0.25/0.09346 =2.675 M

Molality = Number of moles of solute/mass of sovent in kg

Molality = 0.25 mol /0.090 kg = 2.778 mol kg^-1

Ans: The molarity of solution is 2.675mol L^-1 or 2.675 M, the molality of solution is 2.778 mol kg^-1 or 2.778 m, the mole fraction of NaOH is 0. 0476

Example – 09:

A solution of glucose in water is labelled as 10 % (w/w). Calculate a) molality and b) molarity of the solution. Given the density of the solution is 1.20 g mL^-1 and molar mass of glucose is 180 g mol^-1.

Given: density of the solution = 1.20 g cm^-3, % mass of glucose = 10 %, molar mass of glucose is 180 g mol^-1.

To Find: molarity =? and molality =?

Solution:

Consider 100 g of solution

Mass of glucose = 10 g and mass of H₂O = 100 – 10 g = 90 g = 0.090 kg

Molecular mass of water (H₂O) = 1 g x 2 + 16 g x 1 = 18 g mol^-1

Molecular mass glucose = 180 g mol^-1

Number of moles of water = n_A = 90 g/ 18 g = 5 mol

Number of moles of glucose = n_B = 10 g/ 180 g = 0.0556 mol

Density of solution = 1.20 g cm^-3

Volume of solution = Mass of solution / density = 100 g /1.20 g cm^-3 = 83.33 cm³ = 83.33 mL = 0.08333 L

Molarity of solution = Number of moles of the solute/volume of solution in L = 0.0556/0.08333 =0.6672 M

Molality = Number of moles of solute/mass of sovent in kg

Molality = 0.0556 mol /0.090 kg = 0.6178 mol kg^-1

Ans: The molarity of solution is 0.6672 mol L^-1 or 0.6672 M, the molality of solution is 0.6178 mol kg^-1 or 0.6178 m,

Example – 10:

Battery acid 4.22 M aqueous H₂SO₄ solution, and has density 1.21 g cm^-3. What is the molality of H₂SO₄. Given H = 1, S = 32, O = 16

Given: density of the solution = 1.21 g cm^-3, Molarity of solution = 4.22 M.

To Find: molality =?

Solution:

Let us consider 1 L of solution

Molecular mass H₂SO₄ = 1 g x 2 + 32 g x 1 + 16g  x 4 = 98 g mol^-1

Molarity of solution = Number of moles of the solute/volume of solution in L

Number of moles of solute = Molarity of solution x volume of solution in L = 4.22 x 1 = 4.22

Density of solution = 1.21 g cm^-3 = 1.21 g/mL = 1.21 x 10³ g/L = 1.21 kg/L

Mass of solution = Volume of solution x density = 1 L x 1.21 kg/L = 1.21 kg

Mass of solute (H₂SO₄) = Number of moles x molecular mass = 4.22 x 98

Mass of solute (H₂SO₄) = 413.56 g = 0.41356 kg

Mass of solvent = mass of solution – mass of solute = 1.21 – 0.41356 = 0.79644 kg

Molality = Number of moles of solute/mass of sovent in kg

Molality = 4.22 mol /0.79644 kg = 5.298 mol kg^-1

Ans: Molality of solution is 5.298 mol kg^-1 or 5.298 m

Example – 11:

The density of 5.35 M H₂SO₄ solution is 1.22 g cm^-3. What is molality of a solution?

Given: density of the solution = 1.22 g cm^-3, Molarity of solution = 5.35 M.

To Find: molality =?

Solution:

Let us consider 1 L of solution

Molecular mass H₂SO₄ = 1 g x 2 + 32 g x 1 + 16g x 4 = 98 g mol^-1

Molarity of solution = Number of moles of the solute/volume of solution in L

Number of moles of solute = Molarity of solution x volume of solution in L = 5.35 x 1 = 5.35

Density of solution = 1.22 g cm^-3 = 1.22 g/mL = 1.22 x 10³ g/L = 1.22 kg/L

Mass of solution = Volume of solution x density = 1 L x 1.22 kg/L = 1.22 kg

Mass of solute (H₂SO₄) = Number of moles x molecular mass = 5.35 x 98

Mass of solute (H₂SO₄) = 524.3 g = 0.5243 kg

Mass of solvent = mass of solution – mass of solute = 1.22 – 0.5243 = 0.6957 kg

Molality = Number of moles of solute/mass of sovent in kg

Molality = 5.35 mol /0.6957 kg = 7.690 mol kg^-1

Ans: Molality of solution is 7.690 mol kg^-1 or 7.690 m

Example – 12:

Calculate the mole fraction of solute in its 2 molal aqueous solution.

Given: molality = 2 molal

To Find: Mole fraction =?

Solution:

Molecular mass of water (H₂O) = 1 g x 2 + 16 g x 1 = 18 g mol^-1

Molality of solution = 2 molal = 2 mol mol kg^-1

The number of moles of solute = 2

The mass of solvent (water) = 1 kg = 1000 g

Number of moles of solvent (water) = 1000/16 = 55.55

Mole fraction of solute = 2/(2 + 55.55) = 2/57.55 = 0.03475

Ans: Mole fraction of solute is 0.0345

Related Topics

Solutions and Their Colligative Properties

• Solutions and Their Types
• Solutions of Solids and Liquids
• Concentration of Solution
• Numerical Problems on Percentage by Mass and Volume
• Numerical Problems on Mole Fraction
• Numerical Problems on Molarity
• Short Cuts For Above Numerical Problems
• Solutions of Gases in Liquid
• Ideal and Non-ideal Solutions
• Lowering of Vapour Pressure
• Numerical Problems on Lowering of Vapour Pressure
• Elevation in Boiling Point and Depression in Freezing Point
• Osmosis and Osmotic Pressure

For More Topics of Chemistry Click Here

The post Numerical Problems on Molality appeared first on The Fact Factor.

In this article, we shall study numerical problems to calculate the molarity of a given solution.

Example – 01:

A solution of NaOH (molar mass 40 g mol^-1) was prepared by dissolving 1.6 g of NaOH in 500 cm³ of water. Calculate molarity of the NaOH solution.

Given: Mass of NaOH = 1.6 g, molar mass of NaOH = 40 g mol^-1, volume of water = 500 cm³ = 500 mL = 0.5 L

To Find: Molarity =?

Solution:

Number of moles = Given mass/ Molecular mass = 1.6 g/40 g mol^-1 = 0.04 mol

Molarity = Number of moles of solute/Volume of solution in L

Molarity = 0.04 mol /0.5 L = 0.08 mol L^-1

Ans: The molarity of NaOH solution is 0.08 mol L^-1 or 0.08 M

Example – 02:

Calculate molarity of 4 g caustic soda dissolved in 200 mL of solution.

Given: Mass of solute (caustic soda) = 4 g, volume of solution = 200 mL = 0.2 L

To Find: Molarity of the solution =?

Solution:

Molar mass of caustic solute (caustic soda NaOH) = 23 g x1 + 16 g x 1 + 1 g x 1 = (23 + 16 + 1) g = 40 g

Number of moles of caustic solute (caustic soda) = given mass/molecular mass = 4 g/ 40 g = 0.1

Molarity of solution = Number of moles of the solute/volume of solution in L = 0.1/0.2 = 0.5 M

Ans: The molarity of caustic soa solution is 0.5 mol L^-1 or 0.5 M

Example – 03:

Calculate molarity of 5.3 g anhydrous sodium carbonate dissolved in 100 mL of solution.

Given: Mass of solute (sodium carbonate) = 5.3 g, volume of solution = 100 mL = 0.1 L

To Find: Molarity of the solution =?

Solution:

Molar mass of (Na₂CO₃) = 23 g x 2 + 12 g x 1 + 16 g x 3 = (46 + 12 + 48) g = 106 g

Number of moles of (Na₂CO₃) = given mass/molecular mass = 5.3 g/ 106 g = 0.05

Molarity of solution = Number of moles of the solute/volume of solution in L = 0.05/0.1 = 0.5 M

Ans: The molarity of sodium carbonate solution is 0.5 mol L^-1 or 0.5 M

Example – 04:

Calculate molarity of 0.365 g pure HCl gas dissolved in 50 mL of solution.

Given: Mass of solute (HCl) = 0.365 g, volume of solution = 50 mL = 0.05 L

To Find: Molarity of the solution =?

Solution:

Molar mass of HCl = 1 g x 1 + 35.5 g x 1 = (1 + 35.5) g = 36.5 g

Number of moles of HCl = given mass/molecular mass = 0.365 g/ 36.5 g = 0.01

Molarity of solution = Number of moles of the solute/volume of solution in L = 0.01/0.05 = 0.2 M

Ans: The molarity of HCl solution is 0.2 mol L^-1 or 0.2 M

Example – 05:

Calculate molarity of 5.85 g NaCl dissolved in 200 mL of solution.

Given: Mass of solute (NaCl) = 5.85 g, volume of solution = 200 mL = 0.2 L

To Find: Molarity of the solution =?

Solution:

Molar mass of NaCl = 23 g x 1 + 35.5 g x 1 = (23 + 35.5) g = 58.5 g

Number of moles of NaCl = given mass/molecular mass = 5.85 g/ 58.5 g = 0.1

Molarity of solution = Number of moles of the solute/volume of solution in L = 0.1/0.2 = 0.5 M

Ans: The molarity of NaCl solution is 0.5 mol L^-1 or 0.5 M

Example – 06:

Calculate molarity of 20.6 g NaBr dissolved in 500 mL of solution.

Given: Mass of solute (NaCl) = 20.6 g, volume of solution = 500 mL = 0.5 L

To Find: Molarity of the solution =?

Solution:

Molar mass of NaBr = 23 g x 1 + 80 g x 1 = (23 + 80) g = 103 g

Number of moles of NaBr = given mass/molecular mass = 20.6 g/ 103 g = 0.2

Molarity of solution = Number of moles of the solute/volume of solution in L = 0.2/0.5 = 0.4 M

Ans: The molarity of NaBr solution is 0.4 mol L^-1 or 0.4 M

Example – 07:

Calculate molarity of pure water if its density is 1 g/mL.

Given: Density of water = 1 g/mL

To Find: Molarity of pure water =?

Solution:

Let us consider 1000 mL of water

Mass of water = volume x density = 1000 mL x 1 g/mL = 1000 g

Molar mass of water = 1 g x 2 + 16 g x 1 = (2 + 16) g = 18 g

Number of moles of water = given mass/molecular mass = 1000/ 18 g = 55.5

Molarity of pure water = Number of moles of the solute/volume of solution in L = 55.55/1 = 55.55 M

Ans: The molarity of pure water is 55.55 mol L^-1 or 55.5 M

Example – 08:

Calculate the quantity of anhydrous sodium carbonate required to produce 250 mL decimolar solution.

Given: volume of solution = 250 mL = 0.25 L, molarity = decimolar = M/10 = 0.1 M

To Find: Mass of anhydrous sodium carbonate =?

Solution:

Molarity = Number of moles of the solute/volume of solution in L

0.1 = Number of moles of the solute/0.25

Number of moles of the solute = 0.1 x 0.25 = 0.025 mol

Molar mass of (Na₂CO₃) = 23 g x 2 + 12 g x 1 + 16 g x 3 = (46 + 12 + 48) g = 106 g

Number of moles of = given mass/molecular mass

Mass of Na₂CO₃ = Number of moles x molecular mass

Mass of Na₂CO₃ = 106 x 0.025 = 2.65 g

Ans: The quantity of sodium carbonate required is 2.65 g

Example – 09:

Sulphuric acid is 95.8 % by mass. Calculate molarity and mole fraction of H₂SO₄ of density 1.91 g cm^-3. Given H = 1, S = 32, O = 16.

Given: % by mass = 95.8 %, Density of solution = 1.91 g cm^-3

To Find: Mole fraction =? Molarity =?

Solution:

Consider 100 g of solution

Mass of H₂SO₄ = 95.8 g and mass of H₂O = 100 – 95.8 g = 4.2 g

Molecular mass of water (H₂O) = 1 g x 2 + 16 g x 1 = 18 g mol^-1

Molecular mass H₂SO₄ = 1 g x 2 + 32 g x 1 + 16g  x 4 = 98 g mol^-1

Number of moles of water = n_A = 4.2 g/ 18 g = 0.2333 mol

Number of moles of H₂SO₄ = n_B = 95.8 g/ 98 g = 0.9776 mol

Total number of moles = n_A + n_B + n_C = 0.2333 + 0.9776 = 1.2109

Mole fraction of H₂SO₄ = x_B = n_B/(n_A +n_B) = 0.9776/1.2109 = 0.8073

Density of solution = 1.91 g cm^-3

Volume of solution = Mass of solution / density = 100 g /1.91 g cm^-3 = 52.36 cm³ = 52.36 mL = 0.05236 L

Molarity of solution = Number of moles of the solute/volume of solution in L = 0.9776/0.05236 = 18.67 M

Ans: The mole fraction of H₂SO₄ is 0.8073 and molarity of solution is 18.67 mol L^-1 or 18.67 M

Example – 10:

Commercially available concentrated hydrochloric acid is an aqueous solution containing 38% HCl gas by mass. If its density is 1.1 g cm^-3, calculate molarity of HCl solution and also calculate the mole fraction of HCl and H₂O.

Given: % by mass = 38 %, Density of solution = 1.1 g cm^-3

To Find: Mole fraction =? Molarity =?

Solution:

Consider 100 g of solution

Mass of HCl = 38 g and mass of H₂O = 100 – 38 g = 62 g

Molecular mass of water (H₂O) = 1 g x 2 + 16 g x 1 = 18 g mol^-1

Molecular mass HCl = 1 g x 1 + 35.5 g x 1 = 36.5 g mol^-1

Number of moles of water = n_A = 62 g/ 18 g = 3.444 mol

Number of moles of HCl = n_B = 38 g/ 36.5 g = 1.041 mol

Total number of moles = n_A + n_B + n_C = 3.444 + 1.041 = 4.485

Mole fraction of HCl = x_B = n_B/(n_A +n_B) = 1.041/4.485 = 0.2321

Mole fraction of H₂O = x_A = n_A/(n_A +n_B) = 3.444/4.485 = 0.7679

Density of solution = 1.1 g cm^-3

Volume of solution = Mass of solution / density = 100 g /1.1 g cm^-3 = 90.91 cm³ = 90.91 mL = 0.09091 L

Molarity of solution = Number of moles of the solute/volume of solution in L = 1.041/0.09091 = 11.45 M

Ans:  Molarity of solution is 11.45 mol L^-1 or 11.45 M, the molefraction of HCl is 0.2321 and that of H₂O is 0.7679

Example – 11:

Commercially available concentrated hydrochloric acid is an aqueous solution containing 40% HCl gas by mass. If its density is 1.2 g cm^-3, calculate molarity of HCl solution.

Given: % by mass = 40 %, Density of solution = 1.2 g cm^-3

To Find: Molarity =?

Solution:

Consider 100 g of solution

Mass of HCl = 40 g and mass of H₂O = 100 – 40 g = 60 g

Molecular mass HCl = 1 g x 1 + 35.5 g x 1 = 36.5 g mol^-1

Number of moles of HCl = n_B = 40 g/ 36.5 g = 1.096 mol

Density of solution = 1.2 g cm^-3

Volume of solution = Mass of solution / density = 100 g /1.2 g cm^-3 = 83.33 cm³ = 83.33 mL = 0.08333 L

Molarity of solution = Number of moles of the solute/volume of solution in L = 1.096/0.08333 = 13.15 M

Ans:  Molarity of solution is 13.15 mol L^-1 or 13.15 M

Example – 12:

Calculate molarity of a solution containing 50 g of NaCl in 500 g of solution and having density 0.936 g/cm³.

Given: Mass of solute (NaCl) = 50 g, mass of solution = 500 g, density of solution = d = 0.936 g/cm³.

To Find: Molarity of solution = M =?

Molar mass of NaCl = 23 g x 1 + 35.5 g x 1 = (23 + 35.5) g = 58.5 g

Number of moles of NaCl = given mass/molecular mass = 50 g/ 58.5 g = 0.8547

Density of solution = 0.936 g cm^-3

Volume of solution = Mass of solution / density = 500 g /0.936 g cm^-3 = 534.2 cm³ = 534.2 mL = 0.5342 L

Molarity of solution = Number of moles of the solute/volume of solution in L = 0.8547/0.5342 = 1.6 M

Ans: The molarity of NaCl solution is 1.6 mol L^-1 or 1.6 M

Related Topics

Solutions and Their Colligative Properties

• Solutions and Their Types
• Solutions of Solids and Liquids
• Concentration of Solution
• Numerical Problems on Percentage by Mass and Volume
• Numerical Problems on Mole Fraction
• Numerical Problems on Molality
• Short Cuts For Above Numerical Problems
• Solutions of Gases in Liquid
• Ideal and Non-ideal Solutions
• Lowering of Vapour Pressure
• Numerical Problems on Lowering of Vapour Pressure
• Elevation in Boiling Point and Depression in Freezing Point
• Osmosis and Osmotic Pressure

For More Topics of Chemistry Click Here

The post Numerical Problems on Molarity appeared first on The Fact Factor.

The concentration of a solution is the measure of the composition of a solution. For a given solution, the amount of solute dissolved in a unit volume of solution (or a unit volume of solvent) is called the concentration of the solution. It can be expressed either qualitatively or quantitatively. For example, qualitatively we can say that the solution is dilute (i.e., relatively very small quantity of solute) or it is concentrated (i.e., relatively very large quantity of solute). But in practice, it is not useful hence it is not used in chemistry. The quantitative description method gives an exact concentration of the solution and hence its concentration can be compared with the concentration of other solutions.

Methods of Expressing Concentration of the Solution Quantitatively:

Percentage by Mass or Mass Percentage (w/w):

This method is used for a solid in a liquid solution. The mass of solute in gram dissolved in the solvent to form 100 grams of the solution is called percentage by mass. The ratio of the mass of solute to the mass of the solution is called a mass fraction.

For example, if a solution is described by 10% glucose in water by mass, it means that 10 g of glucose is dissolved in 90 g of water resulting in a 100 g solution.

Concentration described by mass percentage is commonly used in industrial chemical applications. For example, a commercial bleaching solution contains a 3.62 mass percentage of sodium hypochlorite in water.

Percentage by Volume (V/V):

This method is used for liquid in a liquid solution. For example, a 10% ethanol solution in water means that 10 mL of ethanol is dissolved in 90 mL water such that the total volume of the solution is 100 mL.

Percentage by Mass by Volume (w/V):

It is the mass of solute dissolved in 100 mL of the solution. This method is commonly used in medicine and pharmacy.

Parts per million:

When a solute is present in trace quantities, it is convenient to express concentration in parts per million (ppm) and is defined as:

As in the case of percentage, concentration in parts per million can also be expressed as mass to mass, volume to volume and mass to volume.

Example: A litre of seawater (which weighs 1030 g) contains about 6 × 10^–3 g of dissolved oxygen (O₂). Such a small concentration is also expressed as 5.8 g per 10⁶ g (5.8 ppm) of seawater. The concentration of pollutants in water or atmosphere is often expressed in terms of ¼ g mL^–1 or ppm.

Strength or Concentration (Grams per litre):

It is defined as the amount of the solute in gram present in the one litre of the solution.

Mole Fraction:

The mole fraction of any component of a solution is defined as the ratio of the number of moles of that component present in the solution to the total number of moles of all components of the solution.

It is to be noted that the sum of the mole fraction of the solute and mole fraction of liquid is 1. The concept of mole fraction is very useful in relating some physical properties of solutions, such as vapour pressure with the concentration of the solution and quite useful in describing the calculations involving gas mixtures. Mole fraction is independent of temperature

Molarity (Molar Concentration):

Molarity (M) is defined as a number of moles of solute dissolved in one litre (or one cubic decimetre) of the solution. The unit of molarity is mol L^-1 0r mol dm^-3 or M.

Number of moles of a substance can be found using the formula

Molarity changes with temperature because volume changes with temperature.

Molarity can be expressed as

• Decimolar = M/10 (0.1 M)
• Semimolar = M/2 (0.5 M)
• Pentimolar = M/5 (0.2 M)
• Centimolar = M/100 (0.01 M)
• milimolar = M/1000 (0.001 M).

Molality:

Molality (m) is defined as a number of moles of solute expressed in kg dissolved in one kg of solvent, Molality has no unit.

Molality is a better way of expressing concentration than molarity because there is no term of volume of solvent is involved. The volume of the solvent depends on the temperature of the solvent. Thus there is no effect of the change of temperature on the molality.

Molality is related to solubility as

Normality:

Normality (N) is defined as gram-equivalent of solute dissolved in one litre (or one cubic decimetre) of the solution, Unit of molarity is N.

A solution having normality equal to unity is called a normal solution.

Decinormal = N/10 (0.1 N), seminormal = N/2  (0.5 N)

Normality × equivalent mass = strength of solution in g/L.

Formality:

Formality is the number of formula mass in gram present per litre of a solution.

If the formula mass of solute is equal to its molar mass, then the formality is equal to molarity. The formality of a solution depends on temperature. This concept is used in the case of ionic substances.

A mole of an ionic compound is called formole and its molarity is called formality. Thus, the formality of a solution may be defined as a number of moles of ionic solute present in one litre of the solution.

The Relation Between Mole Fraction and Molality:

The mole fraction of any component of a solution is defined as the ratio of the number of moles of that component present in the solution to the total number of moles of all components of the solution.

Let us consider a binary solution components solvent (A) and solute (B).

Let x_A = Mole fraction of solvent

x_B = Mole fraction of solute

n_A = Number of moles of solvent

n_B = Number of moles of solute

W_A = Mass of solvent

W_B = Mass of solute

M_A = Molar mass of solvent

M_B = Molar mass of solute

Molarity of Dilution:

Molarity of Mixing:

Relation Between Molarity and Molality:

The density of a solution is in g/mL

Relation Between Molarity and Mole Fraction:

Let x_A = Mole fraction of solvent

x_B = Mole fraction of solute

n_A = Number of moles of solvent

n_B = Number of moles of solute

M = molarity of solution

d = Density of solution

M_A = Molar mass of solvent

M_B = Molar mass of solute

Mass of solution = n_AM_A  +  n_BM_B

Volume os solution = Mass of solution/density of solution

Volume os solution = (n_AM_A  +  n_BM_B)/d

The density of a solution is in g/mL

If density is in g/litre then the molarity is given as

Relation Between Normality and Molarity:

The density of a solution is in g/mL and x is the percentage of solute by mass

Related Topics

Solutions and Their Colligative Properties

• Solutions and Their Types
• Solutions of Solids and Liquids
• Numerical Problems on Percentage by Mass and Volume
• Numerical Problems on Mole Fraction
• Numerical Problems on Molarity
• Numerical Problems on Molality
• Short Cuts For Above Numerical Problems
• Solutions of Gases in Liquid
• Ideal and Non-ideal Solutions
• Lowering of Vapour Pressure
• Numerical Problems on Lowering of Vapour Pressure
• Elevation in Boiling Point and Depression in Freezing Point
• Osmosis and Osmotic Pressure

For More Topics of Chemistry Click Here

The post Concentration of Solution appeared first on The Fact Factor.

In this article, we shall study to solve problems to calculate mole fraction of solute and solvent.

Example – 01:

23 g of ethyl alcohol (molar mass 46 g mol^-1) is dissolved in 54 g of water (molar mass 18 g mol^-1). Calculate the mole fraction of ethyl alcohol and water in solution.

Given: Mass of solute = W_B = 23 g, Molar mass of solute = M_B = 46 g mol^-1, mass of solvent = W_A = 54 g, Molar mass of solvent = M_A = 18 g mol^-1,

To Find: Mole fractions x_B =? x_A = ?

Solution:

Number of moles of solute (ethyl alcohol) = n_B = 23 g/ 46 g mol^-1 = 0.5 mol

Number of moles of solvent (water) = n_A = 54 g/ 18 g mol^-1 = 3 mol

Total number of moles = n_A + n_B = 0.5 + 3 = 3.5 mol

Mole fraction of solute (ethyl alcohol) = x_B = n_B/(n_A + n_B) = 0.5/3.5 = 0.1429

Mole fraction of solvent (water) = x_A = n_A/(n_A + n_B) = 3/3.5 = 0.8571

Ans: Mole fraction of solute (ethyl alcohol) = 0.1429 and mole fraction of solvent (water) = 0.8571

Example – 02:

4.6 cm³ of methyl alcohol is dissolved in 25.2 g of water. Calculate a) percentage by mass of methyl alcohol b) mole fraction of methyl alcohol and water. Given density of methyl alcohol = 0.7952 g cm^-3, and C = 12, H = 1 and O = 16.

Given: Volume of solute (methyl alcohol) = 4.6 cm³, mass of solvent (water) = 25.2 g, density of methyl alcohol = d = 0.7952 g cm^-3,

To Find: percentage by mass of methyl alcohol =?  Mole fraction of methyl alcohol and water =?

Solution:

Mass of methyl alcohol = Volume x density = 4.6 cm³ x 0.7952 g cm^-3 = 3.658 g

Mass of solution = Mass of solute + Mass of solvent = 3.658 g + 25.2 g = 28.858 g

Percentage by mass = (Mass of solute/Mass of solution) x 100

∴ Percentage by mass of urea = (3.658/28.858) x 100 = 12.68%

Molecular mass of methyl alcohol (CH₃OH) = 12 g x 1 + 1 g x 4 + 16g  x 1 = 12 + 4 + 16 = 32 g mol^-1

Number of moles of solute (methyl alcohol) = given mass/ molecular mass

Number of moles of solute (methyl alcohol) = n_B = 3.658 g/ 32 g = 0.1143 mol

Molecular mass of water (H₂O) = n_A = 1 g x 2 + 16g x 1 = 2 + 16 = 18 g mol^-1

Number of moles of solvent (water) = given mass/ molecular mass

Number of moles of solvent (water) = n_B = 25.2 g/ 18 g = 1.4 mol

Total number of moles = n_A + n_B = 0.1143 + 1.4 = 1.5143 mol

Mole fraction of solute (methyl alcohol) = x_B = n_B/(n_A + n_B) = 0.1143/1.5143 = 0.0755

Mole fraction of solvent (water) = x_A = n_A/(n_A + n_B) = 1.2/1.5143 = 0.9245

Ans:  The percentage by mass of methyl alcohol is 12.68% and mole fraction of methyl alcohol is 0.0755 and that of water is 0.9245

Example – 03:

Find the mole fraction of HCl in a solution of HCl containing 24.8 % of HCl by mass. Given H = 1, Cl = 35.5

Given: Percentage by mass = 24.8%

To Find: Mole fraction of HCl =?

Solution:

Percentage by mass of HCl = 24.8%

Let us consider 100 g of HCl solution

Mass of solute (HCl) = 24.8 g

Mass of solvent (water) = 100 – 24.8 = 75.2 g

The molecular mass of HCl = 35.5 g x 1 + 1 g x 1 = 36.5 g

Number of moles of solute (HCl) = given mass/ molecular mass

Number of moles of solute (HCl) = n_B = 24.8 g/ 36.5 g = 0.6795 mol

The molecular mass of water = 1 g x 2 + 16 g x 1 = 18 g

Number of moles of solvent (water) = given mass/ molecular mass

Number of moles of solvent (water) = n_A = 75.2 g/ 18 g = 4.178 mol

Total number of moles = n_A + n_B = 4.178 + 0.6795 = 4.8575 mol

Mole fraction of solute (HCl) = x_B = n_B/(n_A + n_B) = 0.6795/4.8575 = 0.1399

Ans: Mole fraction of HCl = 0.1399

Example – 04:

Calculate the mole fraction of ethylene glycol (C₂H₆O₂) in a solution containing 20% of (C₂H₆O₂) by mass in aqueous solution.

Given: 20% of ethylene glycol (C₂H₆O₂)

To Find: Mole fraction of ethylene glycol (C₂H₆O₂) =?

Solution:

Molecular mass of ethylene glycol (C₂H₆O₂) = 12 g x 2 + 1 g x 6 + 16 g x 2 = 62 g mol^-1

Molecular mass of water (H₂O) = 1 g x 2 + 16 g x 1 = 18 g mol^-1

Let us consider 100 g of ethylene glycol (C₂H₆O₂) solution

Mass of solute (ethylene glycol) = 20 g

Mass of solvent (water) = 100 – 20 = 80 g

Number of moles of solute (ethylene glycol) = n_B = 20 g/ 62 g = 0.3226 mol

Number of moles of solvent (water) = n_B = 80 g/ 18 g = 4.4444 mol

Total number of moles = n_A + n_B = 4.444 + 0.3226 = 4.767 mol

Mole fraction of solute (ethylene glycol) = x_B = n_B/(n_A + n_B) = 0.3226/4.767 = 0.0677

Ans: Mole fraction of ethylene glycol = 0.0677

Example – 05:

Calculate the mole fraction of benzene in a solution containing 30% by mass in carbon tetrachloride.

Given: 30% of benzene in carbon tetrachloride.

To Find: Mole fraction of benzene =?

Solution:

Molecular mass of benzene (C₆H₆) = 12 g x 6 + 1 g x 6 = 78 g mol^-1

Molecular mass of carbon tetrachloride (CCl₄) = 12 g x 1 + 35.5 g x 1 = 154 g mol^-1

Let us consider 100 g of the solution (C₆H₆ + CCl₄)

Mass of solute (ethylene glycol) = 30 g

Mass of solvent (water) = 100 – 30 = 70 g

Number of moles of solute (benzene) = n_B = 30 g/ 78 g = 0.3846 mol

Number of moles of solvent (carbon tetrachloride) = n_B = 70 g/ 154 g = 0.4545 mol

Total number of moles = n_A + n_B = 0.4545 + 0.3846 = 0.8391 mol

Mole fraction of solute (benzene) = x_B = n_B/(n_A + n_B) = 0.3846/0.8391 = 0.4583

Ans: Mole fraction of benzene = 0.4583

Example – 06:

A solution contains 25% water, 25% ethyl alcohol and 50% acetic acid by mass calculate the mole fraction of each component.

Given: 25% water, 25% ethyl alcohol and 50% acetic acid by mass

To Find: mole fraction of each constituent =?

Solution:

Let us consider 100 g of solution

Mass of water = 25 g, Mass of ethyl alcohol = 25 g and mass of acetic acid = 50 g

Molecular mass of water (H₂O) = 1 g x 2 + 16 g x 1 = 18 g mol^-1

Molecular mass of ethyl alcohol (C₂H₅OH) = 12 g x 2 + 1 g x 6 + 16g x 1 = 46 g mol^-1

Molecular mass of acetic acid (CH₃COOH) = 12 g x 2 + 1 g x 4 + 16g x 2 = 60 g mol^-1

Number of moles of water = n_A = 25 g/ 18 g = 1.3889 mol

Number of moles of ethyl alcohol = n_B = 25 g/ 46 g = 0.5435 mol

Number of moles of acetic acid = n_C = 50 g/ 60 g = 0.8333 mol

Total number of moles = n_A + n_B + n_C = 1.3889 + 0.5435 + 0.8333 = 2.7657

Mole fraction of water = x_A = n_A/(n_A +n_B + n_C) = 1.3889/2.7657 = 0.5022

Mole fraction of ethyl alcohol = x_B = n_B/(n_A +n_B + n_C) = 0.5435/2.7657 = 0.1965

Mole fraction of acetic acid = x_C = n_C/(n_A +n_B + n_C) = 0.8333/2.7657 = 0.3013

Related Topics

Solutions and Their Colligative Properties

• Solutions and Their Types
• Solutions of Solids and Liquids
• Concentration of Solution
• Numerical Problems on Percentage by Mass and Volume
• Numerical Problems on Molarity
• Numerical Problems on Molality
• Short Cuts For Above Numerical Problems
• Solutions of Gases in Liquid
• Ideal and Non-ideal Solutions
• Lowering of Vapour Pressure
• Numerical Problems on Lowering of Vapour Pressure
• Elevation in Boiling Point and Depression in Freezing Point
• Osmosis and Osmotic Pressure

For More Topics of Chemistry Click Here

The post Numerical Problems on Mole Fraction appeared first on The Fact Factor.

In this article, we shall learn to calculate percentages by mass and percentage by volume of a solution.

Problems on Percentage by Mass

Example – 01:

6 g of urea was dissolved in 500 g of water. Calculate the percentage by mass of urea in the solution.

Given: Mass of solute (urea) = 6 g, Mass of solvent (water) = 500 g

To Find: Percent by mass =?

Solution:

Mass of solution = Mass of solute + Mass of solvent = 6 g + 500 g = 506 g

Percentage by mass of urea = (Mass of solute/Mass of solution) x 100

= (6/506) x 100 = 1.186%

Example – 02:

34.2 g of glucose is dissolved in 400 g of water. Calculate the percentage by mass of glucose solution.

Given: Mass of solute (glucose) = 34.2 g, Mass of solvent (water) = 400 g

To Find: Percentage by mass =?

Solution:

Mass of solution = Mass of solute + Mass of solvent = 34.2 g + 400 g = 434.2 g

Percentage by mass = (Mass of solute/Mass of solution) x 100

= (34.2/434.2) x 100 = 7.877%

Example – 03:

A solution is prepared by dissolving 15 g of cane sugar in 60 g of water. Calculate the mass percent of each component of the solution.

Given: Mass of solute (cane sugar) = 15 g, Mass of solvent (water) = 60 g

To Find: Mass percent of cane sugar and water =?

Solution:

Mass of solution = mass of solute + mass of solvent = 15 g + 60 g = 75 g

Percentage by mass of solute c(cane sugar) = (Mass of solute/Mass of solution) x 100

Mass percent of solute (cane sugar) = (15 g/75 g) x 100 = 20%

Mass percent of solvent (water) = 100 – 20 = 80%

Example – 04:

Calculate the mass percentage of benzene and carbon tetrachloride if 22 g of benzene is dissolved in 122 g of carbon tetrachloride.

Given: Mass of solute (benzene) = 22 g, Mass of solvent (carbon tetrachloride) = 122 g.

To Find: Mass percentage of benzene and carbon tetrachloride.

Solution:

Mass of solution = mass of solute + mass of solvent

Mass of solution = 22 g + 122 g = 144 g

Percentage by mass = (Mass of solute/Mass of solution) x 100

Percentage of benzene by mass = (22 g/144 g) x 100 = 15.28%

Percentage of carbon tetrachloride by mass = 100 – 15.28 = 84.72%

Example – 05:

A solution is prepared by dissolving a certain amount of solute in 500 g of water. The percentage by mass of a solute in a solution is 2.38. Calculate the mass of solute

Given:  Mass of solvent = 500 g, percentage by mass = 2.38

To Find: Mass of solute =?

Solution:

Let the mass of solute = x g

Mass of solution = Mass of solute + Mass of solvent = x g + 500 g = (x + 500) g

Percentage by mass = (Mass of solute/Mass of solution) x 100

2.38 = (x g/(x + 500) g) x 100

2.38 (x + 500) = 100x

2.38x + 1190 = 100x

1190 = 97.62 x

x = 1190/97.62 = 12.19 g

The mass of solute is 12.19 g

Example – 06:

Calculate the masses of cane sugar and water required to prepare 250 g of 25% cane sugar solution.

Given: 250 g of 25% cane sugar solution

To Find: Masses of cane sugar and water =?

Solution:

Let the mass of cane sugar = x g

Mass of solution = 250 g

Percentage by mass = (Mass of solute/Mass of solution) x 100

25 = (x g/250 g) x 100

25 x 250 g = 100x

6250 g = 100x

x = 6250 g/100 = 62.5 g

Mass of cane sugar = 62.5 g

Mass of water = 250 g – 62.5 g = 187.5 g

Example – 07:

15 g of methyl alcohol is present in 100 mL of solution. If the density of solution is 0.96 g mL^-1. Calculate the mass percentage of methyl alcohol solution.

Given: Mass of solute (methyl alcohol) = 15 g, Volume of solution = V = 100 mL, Density of solution = d = 0.96 g mL^-1.

To Find: mass percentage of methyl alcohol =?

Solution:

Mass of solution = volume x density = 100 mL x 0.96 g mL^-1 = 96 g

Mass percentage = (Mass of solute/Mass of solution) x 100

Mass percentage of benzene = (15 g/96 g) x 100 = 15.625%

Example – 08:

The density of the solution of salt X is 1.15 g mL^-1. 20 mL of the solution when completely evaporated gave a residue of 4.6 g of the salt. Calculate the mass percentage of solute in the solution.

Given: Volume of solution = V = 20 mL, density of solution = d = 1.15 g mL-1, Mass of solute = 4.6 g

To Find: mass percentage of solute in the solution =?

Solution:

Mass of solution = volume x density = 20 mL x 1.15 g mL^-1 = 23 g

Percentage by mass = (Mass of solute/Mass of solution) x 100

Percentage of solute by mass = (4.6 g/23 g) x 100 = 20%

Example – 09:

40% by mass of urea is obtained when 190 g of urea is dissolved in 400 mL of water. Calculate the density of solution.

Given: % by mass of urea solution = 40%, mass of solvent (water) = 400 mL.

To Find: Density of solution =?

Solution:

Percentage by mass = (Mass of solute/Mass of solution) x 100

Mass of solution = (Mass of solute/Percentage by mass) x 100

Mass of solution = (190 g/40) x 100 = 475 g

The volume of solvent (water) = 400 mL = Volume of solution

Density of solution = mass of solution /volume of solution

Density of solution = (475 g) / (400 mL) = 1.19 g mL^-1

Example – 10:

Calculate percent composition in terms of mass of a solution obtained by mixing 300 g of 25% solution of NH₄NO₃ with 400 g of a 40% solution of solute X.

Given: 300 g of 25% solution of NH₄NO₃ mixed with 400 g of a 40% solution of solute X

To Find: percentage composition in terms of mass =?

Solution:

Consider 300 g of 25% solution of NH₄NO₃

Mass of solute in this solution = 25% of 300 g = (25/100) x 300 g = 75 g

Consider 400 g of a 40% solution of solute X

Mass of solute in this solution = 40% of 400 g = (40/100) x 400 g = 160 g

Now let us consider the solution obtained by mixing

Total mass of solute = W_B = 75 g + 160 g = 235 g

Total mass of solution = W_A = 300 g + 400 g = 700 g

Percentage by mass = (Mass of solute/Mass of solution) x 100

Percentage of solute by mass = (235 g/700 g) x 100 = 33.57%

Percentage of solvent by mass = 100 – 33.57 = 66.43%

Example – 11:

Calculate percentage composition in terms of mass of a solution obtained by mixing 100 g of 30% solution of NaOH with 150 g of a 40% solution of NaOH.

Given: 100 g of 30% solution of NaOH mixed with 150 g of a 40% solution of NaOH

To Find: percentage composition in terms of mass =?

Solution:

Consider 100 g of 30% solution of NaOH

Mass of solute in this solution = 30% of 100 g = (30/100) x 100 g = 30 g

Consider 150 g of a 40% solution of NaOH

Mass of solute in this solution = 40% of 150 g = (40/100) x 150 g = 60 g

Now let us consider the solution obtained by mixing

Total mass of solute = W_B = 30 g + 60 g = 90 g

Total mass of solution = W_A = 100 g + 150 g = 250 g

Percentage by mass = (Mass of solute/Mass of solution) x 100

Percentage of solute by mass = (90 g/250 g) x 100 = 36%

Percentage of solvent by mass = 100 – 36 = 64%

Problems on Percentage by Volume:

Example – 12:

12.8 cm³ of benzene is dissolved in 16.8 cm³ of xylene. Calculate percentage by volume of benzene.

Given: Volume of solute = 12.8 cm³, Volume of solvent = 16.8 cm³

To Find: Percentage by volume =?

Solution:

Volume of solution = Volume of solute + Volume of solvent

Volume of solution = 12.8 cm³+ 16.8 cm³ =  29.6 cm³

Percentage by volume = (Volume of solute/Volume of solution) x 100

Percentage of benzene by volume = (12.8 cm³/29.6 cm³) x 100 = 43.24 %

Example – 13:

58 cm³ of ethyl alcohol was dissolved in 400 cm³ of water to form 454 cm³ of a solution of ethyl alcohol. Calculate percentage by volume of ethyl alcohol in water. (12.78 % by volume)

Given: Volume of solute = 58 cm³, Volume of solution = 454 cm³

To Find: Percentage by volume =?

Solution:

Percentage by volume = (Volume of solute/Volume of solution) x 100

Percentage of ethyl alcohol by volume = (58 cm³/454 cm³) x 100 = 12.78%

Related Topics

Solutions and Their Colligative Properties

• Solutions and Their Types
• Solutions of Solids and Liquids
• Concentration of Solution
• Numerical Problems on Mole Fraction
• Numerical Problems on Molarity
• Numerical Problems on Molality
• Short Cuts For Above Numerical Problems
• Solutions of Gases in Liquid
• Ideal and Non-ideal Solutions
• Lowering of Vapour Pressure
• Numerical Problems on Lowering of Vapour Pressure
• Elevation in Boiling Point and Depression in Freezing Point
• Osmosis and Osmotic Pressure

For More Topics of Chemistry Click Here

The post Numerical Problems on Percentage by Mass appeared first on The Fact Factor.