The total pressure observed by mixing two liq | vedclass

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Question

(A) Let the vapour pressure of pure $A$ be $P_{A}^0$ and pure $B$ be $P_{B}^0$.According to Raoult's law,$P_{total} = X_{A}P_{A}^0 + X_{B}P_{B}^0$.For

Vedclass · Accepted Answer

$314$

Vedclass · Answer

(A) Let the vapour pressure of pure $A$ be $P_{A}^0$ and pure $B$ be $P_{B}^0$.According to Raoult's law,$P_{total} = X_{A}P_{A}^0 + X_{B}P_{B}^0$.For the first case: $0.7P_{A}^0 + 0.3P_{B}^0 = 350$ $(i)$For the second case: $0.2P_{A}^0 + 0.8P_{B}^0 = 410$ $(ii)$Multiply equation $(i)$ by $8$ and equation $(ii)$ by $3$:$5.6P_{A}^0 + 2.4P_{B}^0 = 2800$ $(iii)$$0.6P_{A}^0 + 2.4P_{B}^0 = 1230$ $(iv)$Subtracting $(iv)$ from $(iii)$:$5.0P_{A}^0 = 1570$$P_{A}^0 = 314 \ mm \ Hg$.

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