At T (K),the vapour pressures of pure liquids | vedclass

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

Question

(B) Key Idea: The total vapour pressure of the solution is given by $p_{total} = p_A + p_B = \chi_A p_A^{\circ} + \chi_B p_B^{\circ}$.Given:Mole fract

Vedclass · Accepted Answer

$0.294, 0.706$

Vedclass · Answer

(B) Key Idea: The total vapour pressure of the solution is given by $p_{total} = p_A + p_B = \chi_A p_A^{\circ} + \chi_B p_B^{\circ}$.Given:Mole fraction of $A$ in liquid phase,$\chi_A = \frac{2}{2+3} = 0.4$.Mole fraction of $B$ in liquid phase,$\chi_B = \frac{3}{2+3} = 0.6$.$p_A^{\circ} = 100 \ mm$,$p_B^{\circ} = 160 \ mm$.Partial pressure of $A$,$p_A = \chi_A p_A^{\circ} = 0.4 	imes 100 = 40 \ mm$.Partial pressure of $B$,$p_B = \chi_B p_B^{\circ} = 0.6 	imes 160 = 96 \ mm$.Total vapour pressure,$p_{total} = 40 + 96 = 136 \ mm$.Mole fraction of $A$ in vapour state,$y_A = \frac{p_A}{p_{total}} = \frac{40}{136} \approx 0.294$.Mole fraction of $B$ in vapour state,$y_B = \frac{p_B}{p_{total}} = \frac{96}{136} \approx 0.706$.

At $T$ $(K)$,the vapour pressures of pure liquids $A$ and $B$ are $100 \ mm$ and $160 \ mm$ respectively. An ideal solution is formed by mixing $2 \ moles$ of $A$ and $3 \ moles$ of $B$ at the same temperature. The mole fractions of $A$ and $B$ in the vapour state respectively are

Similar Questions

Assertion $(A)$: The vapour pressure of $0.1 \ M$ sugar solution is less than that of $0.1 \ M$ $KCl$ solution.
Reason $(R)$: Lowering of vapour pressure is directly proportional to the number of particles of non-volatile solute present in the solution.
The correct answer is

If the vapor pressure of a solution formed by dissolving $1 \ mol$ of a non-volatile solute in $4 \ mol$ of a solvent is $24.0 \ kPa$,what will be the vapor pressure of the pure solvent in $kPa$?

On adding a solute to a solvent having a vapour pressure of $0.80 \ atm$,the vapour pressure reduces to $0.60 \ atm$. The mole fraction of the solute is:

The vapour pressure of a pure solvent is $0.70 \ atm.$ When a non-volatile solute $A$ is dissolved in it,the vapour pressure of the solution becomes $0.35 \ atm.$ The mole fraction of solute $A$ will be$:-$

Vedclass Test Series

Exam Paper Generator

Online Exam Module