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Numerical Problems on Lowering of Vapour Pressure

Hemant More — Fri, 31 Jan 2020 05:24:53 +0000

Science > Chemistry > Solutions and Their Colligative Properties > Numerical Problems on Lowering of Vapour Pressure

In this article, we shall study to solve problems based on relative lowering of vapour pressure and to calculate the molecular mass of a solute.

Example – 01:

The vapour pressure of a pure liquid at 298K is 4 x 10⁴ N/m². When a non-volatile solute is dissolved the vapour pressure becomes 3.65 x 10⁴ N/m². Calculate relative vapour pressure, lowering of vapour pressure and relative lowering of vapour pressure

Given: Vapour pressure of pure liquid = p^o = 4 x 10⁴ N/m², vapour pressure of solution = p = 3.65 x 10⁴ N/m², temperature = T = 298 K,

To Find: Relative vapour pressure = p/p^o =? Lowering of vapour pressure = p^o – p =? and relative lowering of pressure = (p^o – p)/p^o = ?

Solution:

Relative vapour pressure = p/p^o = (3.65 x 10⁴ N/m²)/(4 x 10⁴ N/m²) = 0.9125

Lowering of vapour pressure = p^o – p = 4 x 10⁴ N/m²– 3.65 x 10⁴ N/m²

Lowering of vapour pressure = 0.35 x 10⁴ N/m²= 3.5 x 10³ N/m²

Relative lowering of pressure = (p^o – p)/p^o = (3.5 x 10³ N/m²)/(4 x 10⁴ N/m²) = 0.0875

Ans: Relative vapour pressure = 0.9125, Lowering of vapour pressure = 3.5 x 10³ N/m², Relative lowering of pressure = 0.0875

Example – 02:

The vapour pressure of a solution containing 13 × 10^-3 kg of solute in 0.1 kg of water at 298 K is 27.371 mm Hg. calculate the molar mass of the solute. Given that the vapour pressure of water at 298 K is 28.065 mm Hg.

Given: mass of solute W₂ = 13 × 10^-3 kg, mass of solvent (water) = W₁ = 0.1 kg, vapour pressure of pure solvent (water) = p^o = 28.065 mm of Hg, vapour pressure of solution = p = 27.371 mm of Hg, temperature = T = 298 K, Molecular mass of solvent (water) = M₁ = 18 g mol^-1

To Find: Molecular mass of solute = M₂ =?

Solution:

For dilute solution relative lowering of vapour pressure is given by

Ans: Molecular mass of solute is 94.63 g mol^-1

Example – 03:

The vapour pressure of pure benzene at a certain temperature is 640 mm Hg. A non-volatile solute of a mass 2.175 × 10^-3 kg is added to 39.0 × 10^-3 kg of benzene. The vapour pressure of a solution is 600 mm Hg. What is the molar mass of the solute? Given C= 12, H = 1.

Given: mass of solute W₂ = 2.175 × 10^-3 kg, mass of solvent = W₁ = 39.0 × 10^-3kg, vapour pressure of pure solvent (benzene) = p^o = 640 mm of Hg, vapour pressure of solution = p = 600 mm of Hg,

To Find: Molecular mass of solute = M₂ =?

Solution:

Molecular mass of solvent (benzene C₆H₆) = M₁ = 12 x 6 + 1 x 6 = 78 g mol^-1

For dilute solution relative lowering of vapour pressure is given by

Ans: Molecular mass of solute is 69.6 g mol^-1

Example – 04:

In an experiment, 18.04 g of mannitol were dissolved in 100 g of water. The vapour pressure of water was lowered by 0.309 mm Hg from 17.535 mm Hg. Calculate the molar mass of mannitol.

Given: mass of solute (mannitol) W₂ = 18.04 g, mass of solvent (water) = W₁ = 100g, vapour pressure of pure solvent (water) = p^o = 17.535 mm of Hg, decrease in vapour pressure of solution = p^o – p = 0.309 mm of Hg, Molecular mass of solvent (water) = M₁ = 18 g mol^-1

To Find: Molecular mass of solute (mannitol) = M₂ =?

Solution:

For dilute solution relative lowering of vapour pressure is given by

Ans: Molecular mass of solute is 184.3 g mol^-1.

Example – 05:

A solution is prepared from 26.2 × 10^-3 kg of an unknown substance and 112.0 × 10^-3 kg acetone at 313 K. The vapour pressure of pure acetone at this temperature is 0.526 atm. Calculate the vapour pressure of solution if the molar mass of a substance is 273.52 × 10^-3 kg mol^-1. Given C = 12, H = 1, O = 16.

Given: mass of solute W₂ = 26.2 × 10^-3 kg, mass of solvent = W₁ = 112.0 × 10^-3kg, vapour pressure of pure solvent (acetone = p^o = 0.526 atm, Molecular mass of solute = M₂ = 273.52 × 10^-3 kg

To Find: vapour pressure of solution = p =?

Solution:

Molecular mass of solvent (acetone (CH₃)₂CO) = M₁ = 12 x 3 + 1 x 6 + 16 x 1

Molecular mass of solvent (acetone (CH₃)₂CO) = M₁ = 58 g mol^-1= 58 × 10^-3 kg

For dilute solution relative lowering of vapour pressure is given by

Ans: Vapour pressure of solution is 0.500 atm.

Example – 06:

The vapour pressure of water at 20 °C is 17 mm Hg. Calculate the vapour pressure of a solution containing 2.8 g of urea (NH₂CONH₂) in 50 g of water.

Given: Temperature of water = 20 °C = 20 + 273 = 293 K, vapour pressure of pure solvent (water) = p^o = 17 mm of Hg, mass of urea (solute)(NH₂CONH₂) = W₂ = 2.8 g, mass of water (solvent) = W₂= 50 g

To Find: Vapour pressure of solution = p =?

Solution:

The molecular mass of solute urea (NH₂CONH₂) = M₂

M₂ = 14 g x 2 + 1 g x 4 + 12 g x 1 + 16 g x 1 = 60 g mol^-1.

The molecular mass of water M₁ = 18 g mol^-1.

Ans: Vapour pressure of solution = 16.714 mm of Hg.

Example – 07:

At 300 K the vapour pressure of water is 1.2 x 10⁴ Pa. 0.8 x 10^-2 kg of oxalic acid (molecular mass = 126) is dissolved in 700 cm³ of water at the same temperature, find the vapour pressure of the solution.

Given: Temperature of water = 300 K, vapour pressure of pure solvent (water) = p^o = 1.2 x 10⁴ Pa, mass of solute (oxalic acid) = W₂ = 0.8 x 10^-2 kg, volume of water (solvent) = 700 cm³, molecular mass of water = M₁ = 18 g mol^-1, molecular mass of solute (oxalic acid) = M₂ = 126 g mol^-1.

To Find: Vapour pressure of solution = p =?

Solution:

Volume of water = 700 cm³

Mass of water = W₁ = 700 cm³ x 1 g cm^-3 = 700 g = 0.7 kg

Ans: Vapour pressure of solution = 1.198 x 10⁴ Pa.

Example – 08:

Calculate the decrease in the vapour pressure when 1.81 g x 10^-2 kg of a solute (molecular mass = 57) is dissolved in 0.1 kg of water. Vapour pressure of water is 1.223 x 10⁴ Pa.

Given: Vapour pressure of pure solvent (water) = p^o = 1.223 x 10⁴ Pa, mass of solute = W₂ = 1.81 x 10^-2 kg, mass of water (solvent) = 0.1 kg, molecular mass of water = M₁ = 18 g mol^-1, molecular mass of solute = M₂ = 57 g mol^-1.

To Find: Decrease in vapour pressure of solution = p^o – p =?

Solution:

Ans: Decrease in vapour pressure is 699 Pa

Example – 09:

A solution containing 30 g of a non-volatile solute in exactly 90 g of water has a vapour pressure of 21.85 mm of Hg at 25 ^oC. Further 18 g of water is then added to the solution. The new vapour pressure becomes 22.15 mm of Hg at 25 ^oC. Calculate the molecular mass of the solute and the vapour pressure of water at 25 ^oC.

Given: Temperature of water = 25 ^oC = 25 + 273 = 298 K

For solution 1: Mass of solute = W₂ = 30 g, mass of solvent = W₁= 90 g, solution vapour pressure = p = 21.85 mm of Hg

For solution 2: Mass of solute = W₂ = 30 g, mass of solvent = W₁= 90 g + 18 g = 108 g, solution vapour pressure = p = 22.15 mm of Hg

To Find: Molecular mass of solute = M₂ =? vapour pressure of pure water solution = p^o =?

Solution:

5p^o – 109.25 = 6p^o – 132.9

p^o = 132.9 -109.25 = 23.65 mm of Hg

Substituting in equation (1)

Ans: Molecular mass of solute is 72.83 u and vapour pressure of pure water solution 23.65 mm

Related Topics

Solutions and Their Colligative Properties

For More Topics of Chemistry Click Here

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Lowering of Vapour Pressure of Solution

Hemant More — Fri, 31 Jan 2020 04:55:13 +0000

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

In this article, we shall learn the meaning of colligative properties, colligative properties of solutions, and the lowering of vapour pressure due to the addition of solute in a solvent.

Colligative Properties:

Colligative properties are those properties of dilute solutions that depend only on the number of solute particles (atoms or molecules, ions or aggregates of molecules) in the solution and not on the nature of solute particles. These properties are important because they are used to determine molar masses of non-electrolyte solutes. Similarly, they are related to one another. If one is measured then other properties can be calculated. These properties can be observed clearly if the solution is very dilute, the solute is non-volatile and the solute does not dissociate or associate in the solution. There are four colligative properties of solutions

Lowering of vapour pressure.
Elevation of the boiling point of the
solvent in the solution.
Depression in the freezing point of
the solvent in the solution.
Osmotic pressure

The Concept of Vapour Pressure:

If a container is partially filled with a liquid, a portion of the liquid evaporates to fill the remaining volume of the container with vapour. The molecules of liquid evaporated are in continuous random motion, they collide with the walls of the container and with each other. Thus they create pressure on the walls of the container and on the liquid. At the same time some molecules which have left liquid return back to the liquid, the process is called condensation. After some interval of time, an equilibrium is established between the two phases of substance. At this stage, the rate of evaporation is equal to the rate of condensation.

The pressure exerted by the vapours of the liquid on the surface of the liquid when equilibrium is established between liquid and its vapour is called vapour pressure of the liquid. The temperature at which vapour pressure of the liquid is equal to the external pressure is called boiling temperature at that pressure.

Relation Between Vapour Pressure and Temperature:

Vapour pressure of liquid increases with the increase in the temperature.

Clausius Clapeyron Equation:

Using this relation we can find vapour pressure at some another temperature when its value at some temperature is known.

Lowering of Vapour Pressure:

Liquids at a given temperature vapourize and under equilibrium conditions, the pressure exerted by the vapours of the liquid over the liquid phase is called vapour pressure [Fig (a)].

In a pure liquid, the entire surface is occupied by the molecules of the liquid. If a non-volatile solute is added to a solvent to give a solution [Fig. (b)], the vapour pressure of the solution is solely from the solvent alone. This vapour pressure of the solution at a given temperature is found to be lower than that of the pure solvent at the same temperature.

In the solution, the surface has both solute and solvent molecules; thereby the fraction of the surface covered by the solvent molecules gets reduced. Consequently, the number of solvent molecules escaping from the surface is correspondingly reduced thus, the vapour pressure is also reduced.

The decrease in the vapour pressure of solvent depends on the quantity of non-volatile solute present in the solution, irrespective of its nature.

Reasons for Lowering of Vapour Pressure:

The evaporation of liquid and evaporation of the liquid is a surface phenomenon. It is directly proportional to the surface area available for evaporation. Due to the addition of non-volatile solute the surface area of liquid decreases. It results in a decrease in the rate of evaporation and vapour pressure decreases.

By Graham’s law, the rate of evaporation is inversely proportional to the square root of the density of the liquid. When a solute is added to a solvent, the density of the resulting solution is more than the pure solvent. Hence the rate of evaporation decreases which results in the decrease of the vapour pressure of the solution.

Relative Lowering of Vapour Pressure:

The relative lowering of vapour pressure for the given solution is the ratio of vapour pressure lowering of solvent from the solution to the vapour pressure of the pure solvent.
Mathematically, the relative lowering of vapour pressure is given by

Where,
p₁⁰= Vapour pressure of the pure solvent
Δp= Lowering of vapour pressure
p= Vapour pressure of the solution

Raoult’s Law:

The partial vapour pressure of any volatile component of a solution is the product of vapour pressure of that pure component and the mole fraction of the component in the solution.

Explanation:

Let us consider a solution containing two volatile components say A₁ and A₂, with mole fractions x₁ and x₂ respectively. Let p₁^o and p₂^obe the vapour pressures of the pure components A₁ and A₂ respectively, then by Raoult’s law

p₁ = p₁^ox₁ and p₂ = p₂^o x₂

The total vapour pressure of the solutions of two volatile components is the sum of partial vapour pressures of the two components

p_T = p₁ + p₂

p_T = p₁^ox₁ + p₂^o x₂

But x₁ + x₂ = 1

∴ x₂ = 1 – x₁

∴ p_T = p₁^ox₁ + p₂^o (1 – x₁)

∴ p_T = p₁^ox₁ + p₂^o – p₂^ox₁

∴ p_T = p₂^o + ( p₁^o – p₂^o)x₁

The solution which obeys Raoult’s law over the entire range of concentration is called an ideal solution. If a solution does not obey Raoul’s law are called non-ideal solutions.

Raoult’s Law for a Solution of Non-Volatile Solute:

Let us consider a solution containing two volatile component A₁ and non-volatile component A₂, with mole fractionsx₁ and x₁ respectively.

Let p₁⁰ and p₂⁰be the vapour pressures of the pure components A₁ and A₂ respectively. Now component A₂ is non-volatile, hence it will not contribute to vapour pressure. Thus p₂⁰= 0. We have

p_T = p₂^o + ( p₁^o – p₂^o)x₁

∴ p_T = 0 + ( p₁^o -0)x₁

∴ p_T = p₁^ox₁

Thus vapour pressure of a solution of non-volatile solute is the product of vapour pressure p₁⁰ of pure solvent and mole fraction x₁ of the solvent, which is Raoult’s law.

The equation shows that vapour pressure of the solution p < p₁⁰ , i.e. there is a lowering of the vapour pressure of the solution mathematically, it is given by

Δp = p₁^o – p

∴ Δp = p₁^o – p₁^ox₁

∴ Δp = p₁^o( 1 – x₁)

∴ Δp = p₁^ox₂

In a solution containing several non-volatile solutes, the lowering of the vapour pressure depends on the sum of the mole fraction of different solutes.
Now, the relative lowering of vapour pressure is given by

This relation proves that the lowering of vapour pressure is colligative property because it depends on the concentration of non-volatile solute.

Raoult’s Law as a Special Case of Henry’s Law:

By Raoult’s law, we have

p = p₁^ox …………… (1)

By Henry’s law, we have

p = K_Hx …………… (2)

If we compare the two equations for Raoult’s law and Henry’s law, it can be seen that the partial pressure of the volatile component or gas is directly proportional to its mole fraction of solute in the solution. Only the proportionality constant K_H differs from p₁⁰. Thus, Raoult’s law becomes a special case of Henry’s law in which K_H is equal to p₁⁰.

Relation Between Molar Mass of Solute and Lowering of Vapour Pressure:

Let W₂ g of the solute of molar mass M₂ be dissolved in W₁ g of the solvent of molar mass M₁. The numbers of moles of solvent and solute are given by n₁ = W₁/M₁ and n2 = W₂/M₂ respectively

The mole fraction of solute is given by

Now for dilute solutions n₂ << n₁. Hence n₂ can be neglected.

This is the relation between the molar mass of solute and the lowering of vapour pressure.

Measurement of Lowering of Vapour Pressure:

Barometric Method:

Raoult introduced the liquid or the solution into the Torricellian vacuum of a barometer tube and measured the depression of the mercury level.

This method is neither practicable nor accurate as the lowering of vapour pressure is too small.

Manometric Method:

This method is generally used for aqueous solutions.

The bulb of the apparatus is charged with the liquid or solution. The air in the connecting tube is then removed with a vacuum pump. When the stopcock is closed, the pressure inside is due only to the vapour evaporating from the solution or liquid. The manometric liquid can be mercury or n-butyl phthalate which has a low density and low volatility.

Ostwald and Walker’s Dynamic Method (Gas Saturation Method):

In this method, the relative lowering of vapour pressure can be determined directly. The measurement of the individual vapour pressures of a solution and solvent is thus eliminated.

Apparatus: The apparatus used is shown in the figure. It consists of two sets of bulbs. The first set of three bulbs is filled with the solution to half of their capacity and second set of another three bulbs is filled with the pure solvent. Each set is separately weighed accurately. Both sets are connected to each other and then with the accurately weighed set of guard tubes filled with anhydrous calcium chloride or some other dehydrating agents like P₂O₅, concentrated H₂SO₄etc. The bulbs of the solution and pure solvent are kept in a thermostat maintained at a constant temperature.

Procedure: A current of pure dry air is bubbled through. The air gets saturated with the vapours in each set of bulbs. The air takes up the amount of vapours proportional to the vapour pressure of the solution first and then it takes up more amount of vapours from the solvent which is proportional to the difference in the vapour pressure of the solvent and the vapour pressure of the solution, i.e. p₀ – p. The two sets of bulbs are weighed again. The guard tubes are also weighed.

Calculation:

The loss in the mass in the solution bulbs ∝ p

The loss in the mass in the solvent bulbs ∝ (p₀ – p)

The total loss of the mass in both sets of bulbs ∝ [p_s + (p₀ – p)] ∝ p₀

The total loss in the mass of both sets of bulbs is equal to gain in mass of guard tubes.

Thus measuring quantities on R.H.S. relative lowering of vapour pressure can be found.

Related Topics

Solutions and Their Colligative Properties

For More Topics of Chemistry Click Here

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