#### Example – 01:

- The vapour pressure of a pure liquid at 298K is 4 x 10
^{4}N/m^{2}. When a non-volatile solute is dissolved the vapour pressure becomes 3.65 x 10^{4}N/m^{2}. calculate relative vapour pressure, lowering of vapour pressure and relative lowering of vapour pressure

**Given:**Vapour pressure of pure liquid = p^{o}= 4 x 10^{4}N/m^{2}, vapour pressure of solution = p = 3.65 x 10^{4}N/m^{2}, 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 10^{4} N/m^{2})/(4 x 10^{4} N/m^{2}) = 0.9125

Lowering of vapour pressure = p^{o} – p = 4 x 10^{4} N/m^{2 }– 3.65 x 10^{4} N/m^{2}

Lowering of vapour pressure = 0.35 x 10^{4} N/m^{2 }= 3.5 x 10^{3} N/m^{2}

relative lowering of pressure = (p^{o} – p)/p^{o} = (3.5 x 10^{3} N/m^{2})/(4 x 10^{4} N/m^{2}) = 0.0875

**Ans: ** Relative vapour pressure = 0.9125, Lowering of vapour pressure = 3.5 x 10^{3} N/m^{2}, 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 W_{2}= 13 × 10^{-3}kg, mass of solvent (water) = W_{1}= 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) = M_{1}= 18 g mol^{-1}**To Find:**Molecular mass of solute = M_{2}= ?**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 W_{2}= 2.175 × 10^{-3}kg, mass of solvent = W_{1}= 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 = M_{2}= ?**Solution:**

Molecular mass of solvent (benzene C_{6}H_{6}) = M_{1} = 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) W_{2}= 18.04 g, mass of solvent (water) = W_{1}= 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) = M_{1}= 18 g mol^{-1}**To Find:**Molecular mass of solute (mannitol) = M_{2}= ?**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 W_{2}= 26.2 × 10^{-3}kg, mass of solvent = W_{1}= 112.0 × 10^{-3 }kg, vapour pressure of pure solvent (acetone = p^{o}= 0.526 atm, Molecular mass of solute = M_{2}= 273.52 × 10^{-3}kg**To Find:**vapour pressure of solution = p = ?**Solution:**

Molecular mass of solvent (acetone (CH_{3})_{2}CO) = M_{1} = 12 x 3 + 1 x 6 + 16 x 1

Molecular mass of solvent (acetone (CH_{3})_{2}CO) = M_{1} = 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 vapour pressure of a solution containing 2.8 g of urea (NH
_{2}CONH_{2}) 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)(NH_{2}CONH_{2}) = W_{2}= 2.8 g, mass of water (solvent) = W_{2 }= 50 g - To Find: Vapour pressure of solution = p = ?
**Solution:**

The molecular mass of solute urea (NH_{2}CONH_{2}) = M_{2}

M_{2} = 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 M_{1} = 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 10
^{4}Pa. 0.8 x 10^{-2}kg of oxalic acid (molecular mass = 126) is dissolved in 700 cm^{3}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 10^{4}Pa, mass of solute (oxalic acid) = W_{2}= 0.8 x 10^{-2}kg, volume of water (solvent) = 700 cm^{3}, molecular mass of water = M_{1}= 18 g mol^{-1}, molecular mass of solute (oxalic acid) = M_{2}= 126 g mol^{-1}.**To Find:**Vapour pressure of solution = p =?**Solution:**

Volume of water = 700 cm^{3}

Mass of water = W_{1} = 700 cm^{3} x 1 g cm^{-3} = 700 g = 0.7 kg

**Ans:** Vapour pressure of solution = 1.198 x 10^{4} 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 10^{4}Pa. **Given:**Vapour pressure of pure solvent (water) = p^{o}= 1.223 x 10^{4}Pa, mass of solute = W_{2}= 1.81 x 10^{-2}kg, mass of water (solvent) = 0.1 kg, molecular mass of water = M_{1}= 18 g mol^{-1}, molecular mass of solute = M_{2}= 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 – 10:

- 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
^{o}C. Further 18 g of water is then added to the solution. The new vapour pressure becomes 22.15 mm of Hg at 25^{o}C. Calculate the molecular mass of the solute and the vapour pressure of water at 25^{o}C. **Given:**Temperature of water = 25^{o}C = 25 + 273 = 298 K**For solution 1:**Mass of solute = W_{2}= 30 g, mass of solvent = W_{1}= 90 g, solution vapour pressure = p = 21.85 mm of Hg**For solution 2:**Mass of solute = W_{2}= 30 g, mass of solvent = W_{1}= 90 g + 18 g = 108 g, solution vapour pressure = p = 22.15 mm of Hg**To Find:**Molecular mass of solute = M_{2}=? 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