The vapour pressures of A and B at 25^{circ} | vedclass

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(A) Given: $P_{A}^{\circ} = 90 \ mm \ Hg$,$P_{B}^{\circ} = 15 \ mm \ Hg$ at $25^{\circ} C$. Mole fraction of $A$ in liquid phase,$X_{A} = 0.6$. Mole f

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(A) Given: $P_{A}^{\circ} = 90 \ mm \ Hg$,$P_{B}^{\circ} = 15 \ mm \ Hg$ at $25^{\circ} C$. Mole fraction of $A$ in liquid phase,$X_{A} = 0.6$. Mole fraction of $B$ in liquid phase,$X_{B} = 1 - 0.6 = 0.4$. Total vapour pressure,$P_{T} = X_{A} P_{A}^{\circ} + X_{B} P_{B}^{\circ}$. $P_{T} = (0.6 	imes 90) + (0.4 	imes 15) = 54 + 6 = 60 \ mm \ Hg$. Partial pressure of $B$ in vapour phase,$P_{B} = X_{B} P_{B}^{\circ} = 0.4 	imes 15 = 6 \ mm \ Hg$. Mole fraction of $B$ in vapour phase,$Y_{B} = \frac{P_{B}}{P_{T}} = \frac{6}{60} = 0.1$. Given $Y_{B} = x 	imes 10^{-1}$,so $0.1 = x 	imes 10^{-1}$,which gives $x = 1$.

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