At a given temperature,the vapour pressure of | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The total vapour pressure of a solution of two volatile liquids is given by $P_S = P_A^o X_A + P_B^o X_B$.Since $X_A + X_B = 1$,we have $X_A = 1 -

Vedclass · Accepted Answer

$150, 90$

Vedclass · Answer

(C) The total vapour pressure of a solution of two volatile liquids is given by $P_S = P_A^o X_A + P_B^o X_B$.Since $X_A + X_B = 1$,we have $X_A = 1 - X_B$.Substituting this into the equation: $P_S = P_A^o (1 - X_B) + P_B^o X_B = P_A^o + (P_B^o - P_A^o) X_B$.Comparing this with the given equation $P_S = 150 - 60 X_B$:For pure $A$ $(X_B = 0)$,$P_S = P_A^o = 150$.For pure $B$ $(X_B = 1)$,$P_S = P_B^o = 150 - 60(1) = 90$.Thus,the vapour pressures of pure $A$ and pure $B$ are $150$ and $90$ respectively.

At a given temperature,the vapour pressure of a solution of two volatile liquids $A$ and $B$ is given by the equation $P_S = 150 - 60 X_B$ (where $X_B$ is the mole fraction of $B$). The vapour pressures of pure $A$ and pure $B$ at the same temperature are respectively:

Similar Questions

The vapour pressure of pure water at $25\,^{\circ}C$ is $23.62\, mm$. What will be the vapour pressure of a solution of $1.5\, g$ urea in $50\, g$ water?

$5 \ cm^3$ of acetone is added to $100 \ cm^3$ of water. The vapour pressure of water over the solution:

At a certain temperature,the mole fraction of a solute is $0.15$ and the vapor pressure of the pure solvent is $120 \, torr$. If the solute is non-volatile,what will be the relative lowering in vapor pressure?

Vapour pressure of a pure solvent is $550 \ mm$ of $Hg$. By addition of a non-volatile solute,it decreases to $510 \ mm$ of $Hg$. Calculate the mole fraction of the solute in the solution.

The weight in grams of a non-volatile solute (mol. wt. $60$) to be dissolved in $90 \ g$ of water to produce a relative lowering of vapour pressure of $0.02$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module