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 :

Question

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 : (A) 10g (B) 11.4g (C) 9.8g (D) 12.8g

Tardigrade Admin · Accepted Answer

Using Raoult's Law, (Po - Ps/Ps)=(W/m)× (M/W) If Po=100mm, then Ps=80mm (100 - 80/80)=(w/114)× (114/40) ∴ w=10g

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 :

$10 g$

$11.4 g$

$9.8 g$

$12.8 g$

Using Raoult's Law,
$\frac{P _{o} - P _{s}}{P _{s}} = \frac{W}{m} \times \frac{M}{W}$
If $P_{o} = 100 mm,$ then $P_{s} = 80 mm$
$\frac{100 - 80}{80} = \frac{w}{114} \times \frac{114}{40}$
$∴ w = 10 g$

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