Ncert Exercises & Solutions

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%.

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, M₂ = 40 g mol^-1

Mass of octane, w₁ = 114 g

Molar mass of octane, (C₈H₁₈), M₁ = 8 × 12 + 18 × 1 = 114 g mol⁻¹

Applying the relation,

$\frac{p_{1}^{0} - p_{1}}{p_{1}^{0}} = \frac{w_{2} \times M_{1}}{M_{2} \times w_{1}} \Rightarrow \frac{p_{1}^{0} - 0.8 p_{1}^{0}}{p_{1}^{0}} = \frac{w_{2} \times 114}{40 \times 114} \Rightarrow \frac{0.2 p_{1}^{0}}{p_{1}^{0}} = \frac{w_{2}}{40} \Rightarrow 0.2 = \frac{w_{2}}{40} \Rightarrow w_{2} = 8 g$

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

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