Vapour pressure of chloroform $(C H C l_{3})$ and dichloromethane $(C H_{2} C l_{2})$ at $298 K$ are $200 m m H g$ and $415 m m H g$ respectively.
(i) Calculate the vapour pressure of the solution prepared by mixing $25.5 g$ of $C H C l_{3}$ and $40 g$ of $C H_{2} C l_{2}$ at $290 K$ and
(ii) Mole fraction of each component in solution.

Open in App

Solution

Vapour pressure of pure chloroform = $P_{A}^{o} = 200 m m$
Vapour pressure of pure dichloromethane $P_{B}^{o} = 415 m m H g$
Molar mass $M_{A} = 119.5 g$
Molar mass $M_{B} = 85 g$
(i) mass of $C H C l_{3} = 25.5 g$
number of mole $= \frac{25.5}{119.5} = 0.213$
Mass of $C H_{2} C l_{2} = 40 g$
Number of mole $= \frac{40}{85} = 0.47$
$x_{A} = \frac{0.213}{0.213 + 0.47} = 0.322$
$x_{B} = 1 - 0.322 = 0.688$
$P = P_{A}^{o} X_{A} + P_{B}^{o} x_{B}$
$= 200 \times 0.322 + 415 \times 0.688$
$= 64.4 + 285.52 = 349.92$
(ii) Mole fraction of chlotoform = $0.322$
Mole fraction of $C H_{2} C l_{2} = 0.688$

Suggest Corrections

Similar questions

(a) Vapour pressure of chloroform $(C H C l_{3})$ and dichloromethane $(C H_{2} C l_{2})$ at 298 K are 200 mm Hg and 415 mm Hg respectively. Calculate the vapour pressure of the solution prepared by mixing 25.5 g of CHCl3 and 40 g of $C H_{2} C l_{2} a t 298 K$