The mass of a non-volatile solute of molar mass 40 g mol-1 that should be dissolved in 114 g of octane to lower its vapour pressure by 20 % is

Question

The mass of a non-volatile solute of molar mass 40 g mol-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

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/6a59c8490f8b753833e6d6e95e110389-.png" /> ∵ (p°-p/p)=(w/m) × (M/W) ∴ (100-80/80)=(w/40) × (114/114) or (20/80)=(w/40) or w=(20 × 40/80)=10 g

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

$10 g$

$11.4 g$

$9.8 g$

$12.8 g$

$∵ \frac{p ^{\circ} - p}{p} = \frac{w}{m} \times \frac{M}{W}$

$∴ \frac{100 - 80}{80} = \frac{w}{40} \times \frac{114}{114}$

or $\frac{20}{80} = \frac{w}{40}$

or $w = \frac{20 \times 40}{80} = 10 g$

Q. The mass of a non-volatile solute of molar mass 40gmol−1 that should be dissolved in 114g of octane to lower its vapour pressure by 20% is

10g

11.4g

9.8g

12.8g