Liquids A and B form an ideal solution over t | vedclass

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(D) The vapour pressure of pure liquid $A$ is $P_A^0 = 20 \ Torr$.For the equimolar solution,$X_A = 0.5$ and $X_B = 0.5$. The total vapour pressure is

Vedclass · Accepted Answer

$19$

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(D) The vapour pressure of pure liquid $A$ is $P_A^0 = 20 \ Torr$.For the equimolar solution,$X_A = 0.5$ and $X_B = 0.5$. The total vapour pressure is $P_{	ext{Total}} = P_A^0 X_A + P_B^0 X_B = 45 \ Torr$.Substituting the values: $45 = 20(0.5) + P_B^0(0.5) \implies 45 = 10 + 0.5 P_B^0 \implies 35 = 0.5 P_B^0 \implies P_B^0 = 70 \ Torr$.For the new solution,$P_{	ext{Total}} = 22.5 \ Torr = P_A^0 X_A + P_B^0 X_B$.Since $X_A + X_B = 1$,we have $X_B = 1 - X_A$.$22.5 = 20 X_A + 70(1 - X_A) \implies 22.5 = 20 X_A + 70 - 70 X_A$.$50 X_A = 47.5 \implies X_A = 0.95$.Then $X_B = 1 - 0.95 = 0.05$.Therefore,$\frac{x_A}{x_B} = \frac{0.95}{0.05} = 19$.

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