The vapour pressure of pure liquids A and B a | vedclass

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Question

(A) Given,vapour pressure of pure liquid $A$,$p_{A}^{\circ} = 450 \ mm \ Hg$. Vapour pressure of pure liquid $B$,$p_{B}^{\circ} = 700 \ mm \ Hg$. Tota

Vedclass · Accepted Answer

$\chi_{A}=0.4, \chi_{B}=0.6$

Vedclass · Answer

(A) Given,vapour pressure of pure liquid $A$,$p_{A}^{\circ} = 450 \ mm \ Hg$. Vapour pressure of pure liquid $B$,$p_{B}^{\circ} = 700 \ mm \ Hg$. Total vapour pressure,$p_{	ext{Total}} = 600 \ mm \ Hg$. From Raoult's law,$p_{	ext{Total}} = p_{A}^{\circ}\chi_{A} + p_{B}^{\circ}\chi_{B}$. Since $\chi_{B} = 1 - \chi_{A}$,we have: $600 = 450\chi_{A} + 700(1 - \chi_{A})$ $600 = 450\chi_{A} + 700 - 700\chi_{A}$ $600 = 700 - 250\chi_{A}$ $250\chi_{A} = 100$ $\chi_{A} = \frac{100}{250} = 0.4$ $	herefore \chi_{B} = 1 - 0.4 = 0.6$.

The vapour pressure of pure liquids $A$ and $B$ are $450 \ mm$ and $700 \ mm$ of $Hg$ at $350 \ K$ respectively. If the total vapour pressure of the mixture is $600 \ mm$ of $Hg$,the composition of the mixture in the solution is

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