**Question:**

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%.

**Solution:**

Let the vapour pressure of pure octane be $p_{1}^{0}$.

Then, the vapour pressure of the octane after dissolving the non-volatile solute is $\frac{80}{100} p_{1}^{0}=0.8 p_{1}^{0}$.

Molar mass of solute, *M*2 = 40 g mol−1

Mass of octane, *w*1 = 114 g

Molar mass of octane, (C8H18), *M*1 = 8 × 12 + 18 × 1

= 114 g mol−1

Applying the relation,

$\frac{p_{1}^{0}-p_{1}}{p_{1}^{0}}=\frac{w_{2} \times M_{1}}{M_{2} \times w_{1}}$

$\Rightarrow \frac{p_{1}^{0}-0.8 p_{1}^{0}}{p_{1}^{0}}=\frac{w_{2} \times 114}{40 \times 114}$

$\Rightarrow \frac{0.2 p_{1}^{0}}{p_{1}^{0}}=\frac{w_{2}}{40}$

$\Rightarrow 0.2=\frac{w_{2}}{40}$

$\Rightarrow w_{2}=8 \mathrm{~g}$

Hence, the required mass of the solute is 8 g.

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