Question

The mass of a non-volatile solute of molar mass 40 g $mo{l}^{-1}$ that should be dissolved in 114 g of octane to lower its vapour pressure by 20% is :

• A. 10 g
• B. 11.4 g
• C. 9.8 g
• D. 12.8 g

Answer 1

Answer: (A)

10 g

The mass of a non-volatile solute of molar mass 40 g /mol that should be dissolved in 114 g of octane to lower its vapour pressure by 20% is 10 g.

Let w g of the non-volatile solute is used. Its molar mass is 40 g/mol.

The number of moles of non volatile solute $=\frac{w}{40}=0.025w$ moles.

114 g of octane (molar mass 114 g/mol) corresponds to $\frac{114}{114}=1$ mole.

The mole fraction of non volatile solute is $=\frac{0.025w}{1+0.025w}$. The mole fraction of non volatile solute is proportional to the relative lowering in the vapour pressure.

$\frac{P^0-P}{P^0}=X_B$

$\frac{20}{100}=\frac{0.025w}{1+0.025w}$

$100\left(0.025w\right)=20(1+0.025w)$

$2.5w=20+0.50w$

$2w=20$

$w=10$ g

Hence, the correct option is $\text{A}$