Ncert Exercises & Solutions

Molar mass of water = 18 g mol⁻¹

Number of moles present in 1000 g of water = 55.56 mol
$∴$ Number of moles are present in 1000 g of water= $\frac{1000}{18} = 55.56$ mol

Therefore, mole fraction of the solute in the solution is

$x_{2} = \frac{1}{1 + 55.56} = 0.0177$

It is given that,

Vapour pressure of water, $p_{1}^{0} = 12.3 kPa$

Applying the relation,

$\frac{p_{1}^{0} - p_{1}}{p_{1}^{0}} = x_{2}$

$\Rightarrow \frac{12.3 - p_{1}}{12.3} = 0.0177$

⇒ 12.3 − p₁ = 0.2177

⇒ p₁ = 12.0823

= 12.08 kPa (approximately)

Hence, the vapour pressure of the solution is 12.08 kPa.

2.18 Calculate the mass of a non-volatile solute (molar mass 40 g mol–1) which should be dissolved in 114 g octane to reduce its vapour pressure to 80%.

1 molal solution means 1 mol of the solute is present in 100 g of the solvent (water).

Molar mass of water = 18 g mol⁻¹

Number of moles present in 1000 g of water = 55.56 mol
$∴$ Number of moles are present in 1000 g of water= $\frac{1000}{18} = 55.56$ mol

Therefore, mole fraction of the solute in the solution is

$x_{2} = \frac{1}{1 + 55.56} = 0.0177$

It is given that,

Vapour pressure of water, $p_{1}^{0} = 12.3 kPa$

Applying the relation,

$\frac{p_{1}^{0} - p_{1}}{p_{1}^{0}} = x_{2}$

$\Rightarrow \frac{12.3 - p_{1}}{12.3} = 0.0177$

⇒ 12.3 − p₁ = 0.2177

⇒ p₁ = 12.0823

= 12.08 kPa (approximately)

Hence, the vapour pressure of the solution is 12.08 kPa.

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