Numerical Problems on Lowering of Vapour Pressure

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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^-3 kg, 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^-3 kg, 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

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