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Ideal and Non-Ideal solutions

Hemant More — Fri, 31 Jan 2020 06:06:33 +0000

Science > Chemistry > Solutions and Their Colligative Properties > Ideal and Non-Ideal Solutions

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.

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Short-cut Methods For Calculating Concentration of Solutions

Hemant More — Thu, 30 Jan 2020 14:36:14 +0000

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

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For More Topics of Chemistry Click Here

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Numerical Problems on Molality

Hemant More — Thu, 30 Jan 2020 01:47:16 +0000

Science > Chemistry > Solutions and Their Colligative Properties > Numerical Problems on Molality

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^-1or 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^-1and 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^-1or 0.1325 M, the molality of the solution is 0.1342 mol kg^-1or 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^-1or 3.374 M, the molality of solution is 3.774 mol L^-1or 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^-1or 2.011 M, the molality of solution is 2.206 mol kg^-1or 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^-1or 1.040 M

The molality of solution is 1.067 mol kg^-1or 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^-1or 2.675 M, the molality of solution is 2.778 mol kg^-1or 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^-1or 0.6672 M, the molality of solution is 0.6178 mol kg^-1or 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

For More Topics of Chemistry Click Here

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Numerical Problems on Molarity

Hemant More — Wed, 29 Jan 2020 18:49:26 +0000

Science > Chemistry > Solutions and Their Colligative Properties > Numerical Problems on Molarity

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^-1or 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^-1or 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^-1or 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^-1or 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^-1or 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^-1or 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^-1or 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^-1or 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^-1or 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^-1or 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^-1or 1.6 M

Related Topics

Solutions and Their Colligative Properties

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Concentration of Solution

Hemant More — Wed, 29 Jan 2020 18:28:02 +0000

Science > Chemistry > Solutions and Their Colligative Properties > Concentration of Solution

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:

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

For More Topics of Chemistry Click Here

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Numerical Problems on Mole Fraction

Hemant More — Wed, 29 Jan 2020 18:20:04 +0000

Science > Chemistry > Solutions and Their Colligative Properties > Numerical Problems on Mole Fraction

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

For More Topics of Chemistry Click Here

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

Numerical Problems on Percentage by Mass

Hemant More — Wed, 29 Jan 2020 18:03:12 +0000

Science > Chemistry > Solutions and Their Colligative Properties > Percentage Composition

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 NH₄NO₃mixed 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

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Solid and Liquid Solutions

Hemant More — Wed, 29 Jan 2020 14:21:41 +0000

Science > Chemistry > Solutions and Their Colligative Properties > Solid and Liquid Solutions

In this article, we shall study solid and liquid solutions and the concept of solubility curves.

A solution of Solids in Liquids:

The dissolution of any substance in a liquid to form a solution is governed by the basic principle that the solute-solvent interaction is either similar to or greater than solute-solute and solvent-solvent interaction. Let us consider the dissolution of an ionic solid in water. There is an interionic attraction between the ions of solid. In this case, the cations and anions of the solid get attracted by the opposite ends of water dipoles.

If the ion-dipole interaction forces are stronger than interionic attraction, the ions are pulled out of crystal lattice and they pass into solution. Formed ions are called hydrated ions.

Molecular solids dissolve in the water on account of their capacity to form hydrogen bonds with water molecules.

The energy required to dismantle 1 mol of the crystal lattice is called lattice energy (Δ_LH). The energy released during hydration is called hydration energy (Δ_HydroH). Mathematically,

Δ_SolH = Δ_LH + Δ_HydroH.

Case – I: Δ_HydroH ≥ Δ_LH:Dissolution occurs and Δ_SolH < 0
Case – II: Δ_HydroH marginally less than Δ_LH: Dissolution occurs and Δ_SolH > 0
Case – III: Δ_LH too large to compensate Δ_HydroH: Dissolution does not occur.

Factors Affecting Solubility of a Solids in Liquids:

Nature of solutes: Polar solvent (e.g. water) dissolves polar solute (e.g. NaCl, KCl) and non-polar solvents (CCl₄, CS₂) dissolve non-polar solutes (e.g. I₂, S₈).
Temperature: The solubility of a solid in a liquid is significantly affected by temperature changes. Consider the equilibrium of solute particles dissolving and crystallizing.

Solute + Solvent ⇌ Solution

It is a dynamic equilibrium and follows Le Chatelier,s Principle. If in a nearly saturated solution, the dissolution process is endothermic (Δ_solH > 0), the solubility should increase with the rise in temperature if it is exothermic (Δ_solH < 0) the solubility should decrease.

Pressure: Pressure does not have any significant effect on so the ability of solids in liquids. It is so because solids and liquids are highly incompressible and practically remain unaffected by changes in pressure.

A solution of Solids in Solids:

A solid solution of two or more metals or metals or metals with one or more non-metal is called an alloy.

Types of Solid Solutions:

Substitutional: Solute
atoms occupy the regular lattice sites of the parent metal (solvent).
Substitutional solid solutions can be random (Cu-Ni) or ordered (Cu-Au).
Interstitial: Solute
atoms occupy the interstitial positions (Steel – C solute atoms in Fe).

Examples of Alloys:

Alloy	Composition	Uses
Duralumin	Aluminum, copper, manganese, magnesium	aircraft, boats, railroad cars, and machinery because of its high strength and resistance to corrosion
Aluminium bronze	Aluminum, copper, manganese	heavy duty sleeve bearings, and machine tool ways. Aluminium bronze castings have exceptional corrosion resistance, high strength, toughness and wear resistance and good casting and welding characteristics.
Hardened Lead	10 to 20 % antimony	bearings, bullets and shrapnel
Stainless Steel	Iron, chromium, nickel	Cutlery, surgical instruments, utensils
Spiegeleisen	Iron and 5 to 20 % manganese	Very hard hence used for making rails, safes and heavy machinery
Ferromanganeous	Iron and 70 to 80 % manganese	Very hard hence used for making rails, safes and heavy machinery
Manganin	84 % Cu, 12 % Mn, 4% Ni	electrical measurement instrument due to zero temperature coefficient of resistance.

Amalgams:

Alloys of mercury with other metals are called amalgams. This property of mercury is used for extraction of metals like silver and gold from their ores. These metals are further recovered by distillation of mercury.

A solution of Liquids in Liquids:

Evolution or Absorption of Heat During Formation of Solution:

When one liquid is dissolved in another, two processes should take place simultaneously. In the first, the solvent molecules move apart to accommodate the solute molecules while in the second, the solute molecules separate to get accommodated in the space provided by the solvent. For this process energy and absorbed. At the end of the formation of the solution, the solute and solvent molecules are brought together. This results in the release of energy. The net of the two energies gives the enthalpy of solution.

If there is a strong attraction between the solute and the solvent molecules, the heat released is more than the heat absorbed and more energy is released in the final step. Thus with the concept of energy, there are thee possibilities.

Case – I: There is the absorption of heat (e.g. formation of a
solution of ethyl alcohol and water).
Case – II: There is an evolution of heat (e.g. formation of a solution
of acetone and water)
Case – III: There is neither absorption nor evolution of heat (e.g.
formation of a solution of benzene and carbon tetrachloride)

Effect of Relative Solubility of Solute in Solution:

When two liquids are mixed, there are three possibilities.

Case – I: The two components are almost immiscible. It happens when
one liquid is polar and other in non-polar. e.g. a solution of water and
benzene, a solution of carbon tetrachloride and water, a solution of
benzene and alcohol.
Case – II: The miscibility of the component may be partial. This
happens when the intermolecular attraction of one liquid is different from the
intermolecular attraction of other liquid. e.g. a solution of water and
ether, a solution of phenol and water, a solution of nicotine and
water.
Case – III: The two components may be completely miscible. It happens
when the two liquids have the same nature. Either both are polar (e.g. a
solution of alcohol and water) or both are non-polar (e.g. a solution of
benzene and hexane, a solution of benzene and toluene).

Characteristics of miscible liquids:

They are chemically alike.
Both have the same nature.
Dipole-dipole interaction is stronger.
Molecular sizes of liquids are approximately the same.

Solubility Curves:

Solubility is measured by determining the maximum mass of a solute that can be dissolved in 100 g of a solvent at a given temperature. A solubility curve is a graph giving the relationship between the temperature and solubility of a particular solute. Thus the solubility curves give the idea about what mass of solute will dissolve in 100g (or 100mL) of water over a range of temperatures.

Solutes whose solubility curves move upward with increased temperature are typically solids because the solubility of solids increases with increased temperature. The steeper the incline of a solute, the more soluble the solute is because it doesn’t take as much of a temperature increase to dissolve the substance. This also means that more of the solute can be dissolved versus another substance at the same temperature. From the solubility curve, we can see that the solubility of KNO₃ rises sharply with the increase in temperature. In solutions involving liquids and solids typically more solute can be dissolved at higher temperatures.

Solutes whose solubility curves move downward with increased temperature are typically gases because the solubility of gases decreases with increased temperature.

The flatter the line, the less soluble the solute it, because it takes a larger temperature in order for a solute to dissolve compared to a
steeper inclined substance. The solubility of NaCl almost remains constant.

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Solutions and Their Types

Hemant More — Wed, 29 Jan 2020 13:49:28 +0000

Science > Chemistry > Solutions and Their Colligative Properties > Solutions and Their Types

In this article, we shall study the concept of solutions and their types based on phases, composition, etc.

The Terminology of Solutions:

Solution: A solution is a homogeneous mixture of two or more than two or more components. A solution has a single phase. The constituent particles of solution can not be separated by filtration, settling or centrifugal action.
Solvent: The component of a solution which dissolves the other component in itself is called a solvent. It is the larger component of the solution. For example, a solution of sugar in water is solid in the liquid. Here, sugar is the solute and water is the solvent.
Solute: The component of the solution which dissolves in the solvent is called the solute. The solute is the smaller component of the solution. For example, a solution of sugar in water is solid in the liquid. Here, sugar is the solute and water is the solvent.
Homogeneous Solution: If the composition and properties are uniform throughout the mixture, the solution is called a homogeneous solution.
If the solution consists of two chemical components, the solution is called a binary solution. If it contains three or four chemical components it is called a ternary or quaternary solution.
Soluble Substance: A substance that dissolves in a solvent is said to be soluble in that solvent. A soluble substance is able to dissolve in a solvent because attractive forces between the solvent and solute particles are strong enough to overcome the attractive forces holding the solute particles together.
Insoluble Substance: A substance that does not dissolve in a solvent is called insoluble in that solvent.
Miscible Liquids: Two liquids that are soluble in each other are said to be miscible such as water and vinegar, coffee and cream
Immiscible Liquids: Liquids that are not soluble in each other are immiscible such as vegetable oil and vinegar, petrol and water.
The Process of Dissolving: Solvent particles surround solute particles to form a solution in a process called solvation. If the solvent is water the process is known as hydration. During this process, there is a change in energy observed by an increase or decrease in temperature.
Like dissolve Like: Polar dissolves polar & nonpolar dissolves nonpolar.
Aqueous Solution: Solutions having water as a solvent are called aqueous solutions.

The concept of Solubility:

When a substance (solute) gets dissolved in a liquid (solvent) to form a solution it means solute-solvent interaction is greater than solute-solute. The solubility of a substance is its maximum amount that can be dissolved in a specified amount of solvent. It depends upon the nature of the solute and solvent as well as temperature and pressure.

Let us consider the effect of these factors in a solution of a solid or a gas in a liquid. When a solid solute is added to the solvent, some solute dissolves and its concentration increases in solution. This process is known as dissolution. Some solute particles in solution collide with the solid solute particles and get separated out of solution. This process is known as crystallization. A stage is reached when the two processes occur at the same rate. Under such conditions, the number of solute particles going into the solution will be equal to the solute particles separating out and a state of dynamic equilibrium is reached. At this stage, the concentration of solute in solution will remain constant under the given conditions, i.e., temperature and pressure. A similar process is followed when gases are dissolved in liquid solvents.

Such a solution in which no more solute can be dissolved at the same temperature and pressure is called a saturated solution. An unsaturated solution is one in which more solute can be dissolved at the same temperature. The maximum amount of solute that dissolves completely in a given amount of solvent at constant temperature is called the solubility of the solute.

Factors that affect solubility:

The solubility of a solid in liquid increases with the increase in temperature. Whereas, the solubility of a gas in the liquid decreases with the increase in temperature.
A particular volume of solvent can dissolve some maximum amount of solute. Thus the rate of dissolving decreases with the increase in the amount of solute.
Stirring or agitating solution increases the rate of dissolving. Agitation increases the rate by bringing fresh solvent into contact with more solute
Dissolving occurs at the surface of a solid, by increasing the surface area we can increase the rate of dissolving.

TYpes of Solutions:

On the Basis of Size of Solute Particles:

Depending upon the size of solute particles, the solutions are broadly classified into three types.

Corse Solution: When the size of constituent particles is relatively bigger, then the solution is called a coarse solution e.g. a mixture of salt and sugar.
Colloidal Solution: A colloidal solution is formed when the size of the particles dispersed in the solvent is in the range of 10^-7 cm to 10^-4 cm. The colloidal particles carry a charge of the same nature, which is important for the stabilization of the solution e.g. ferric hydroxide sol
True Solution: A true solution is defined as a homogeneous mixture of two or more substances, the composition of which is not fixed and may be varied within a certain limit. The size of the particles dispersed in the solvent is less than 10^-4 cm. e.g. Solution of common salt in water.

Characteristics of True Solutions

A solution is a
homogeneous mixture
The size of
solute particles in the solutions is extremely small. It is less than 1 nm in
diameter.
The particles of
a solution cannot be seen even with a microscope.
The particles of
a solution pass through the filter paper. Thus filtration cannot separate the
solution.
It is very
stable. The particles of solute present in a solution do not separate out on
keeping.
A true solution
does not scatter light (because its particles are very small).

On the Basis of Phases of Solute and Solvent:

Depending upon the phases of solute and solvent the solutions are broadly classified into three types.

Gaseous Solutions:

Solute phase: solid, solvent phase: gas, Also known as solid in gas solutions e.g. iodine in air
Solute phase: liquid, solvent phase: gas, Also known as a liquid in gas solutions e.g. Chloroform in nitrogen
Solute phase: gas, solvent phase: gas, also known as gas in gas solutions e.g. Mixture of non-reacting gases (O₂+ N₂)

Liquid Solutions:

Solute phase: solid, solvent phase: liquid, Also known as solid in liquid solutions e.g. sugar in water
Solute phase: liquid, solvent phase: liquid, Also known as a liquid in liquid solutions e.g. ethanol in water
Solute phase: gas, solvent phase: liquid, Also known as gas in liquid solutions e.g. carbon dioxide in water

Solid Solutions:

Solute phase: solid, solvent phase: solid, Also known as solid in solid solutions e.g. Alloys like brass, bronze
Solute phase: liquid, solvent phase: solid, Also known as a liquid in solid solutions e.g. Amalgam of mercury with metal
Solute phase: gas, solvent phase: solid, Also known as gas in solid solutions e.g. pumice stone, H₂ gas in palladium

Sometimes solutions of liquid in gas and solid in a gas are not considered as solutions because the mixture may not be homogeneous.

On the Basis of Concentrations:

Saturated solution: A saturated solution is defined as the solution that contains a just amount of dissolved solute necessary for establishing equilibrium between dissolved solute and undissolved solids.
Unsaturated solution: An unsaturated solution is defined as a solution in which more solute can be dissolved at the same temperature.
Supersaturated solution: A supersaturated solution is defined as a solution in which excess solute is dissolved than required for the formation of a saturated solution

Related Topics

Solutions and Their Colligative Properties

For More Topics of Chemistry Click Here

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