The vapour pressures of pure liquids A and B | vedclass

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Question

(A) According to Raoult's Law,the total vapour pressure of the solution is given by $P_{total} = X_A P_A^o + X_B P_B^o$.Given $X_B = 0.5$,therefore $X

Vedclass · Accepted Answer

$500\, mm\, Hg, 0.4, 0.6$

Vedclass · Answer

(A) According to Raoult's Law,the total vapour pressure of the solution is given by $P_{total} = X_A P_A^o + X_B P_B^o$.Given $X_B = 0.5$,therefore $X_A = 1 - 0.5 = 0.5$.$P_{total} = (0.5 	imes 400) + (0.5 	imes 600) = 200 + 300 = 500\, mm\, Hg$.Now,the mole fraction of component $A$ in the vapour phase $(Y_A)$ is calculated as $Y_A = \frac{P_A}{P_{total}} = \frac{X_A P_A^o}{P_{total}} = \frac{0.5 	imes 400}{500} = 0.4$.Similarly,the mole fraction of component $B$ in the vapour phase $(Y_B)$ is $Y_B = 1 - Y_A = 1 - 0.4 = 0.6$.Thus,the values are $500\, mm\, Hg, 0.4, 0.6$.

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