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

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(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

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