At 350 K,the vapour pressure of pure liquids | vedclass

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(C) Given: $P_A^0 = 450 \ mm \ Hg$,$P_B^0 = 700 \ mm \ Hg$,$P_{total} = 600 \ mm \ Hg$. According to Raoult's Law: $P_{total} = P_A^0 x_A + P_B^0 x_B$

Vedclass · Accepted Answer

$0.3, 0.7$

Vedclass · Answer

(C) Given: $P_A^0 = 450 \ mm \ Hg$,$P_B^0 = 700 \ mm \ Hg$,$P_{total} = 600 \ mm \ Hg$. According to Raoult's Law: $P_{total} = P_A^0 x_A + P_B^0 x_B$. Since $x_A + x_B = 1$,we have $x_B = 1 - x_A$. Substituting values: $600 = 450 x_A + 700(1 - x_A)$. $600 = 450 x_A + 700 - 700 x_A$. $250 x_A = 100$,so $x_A = 100 / 250 = 0.4$. Thus,$x_B = 1 - 0.4 = 0.6$. Partial pressures in vapour phase: $P_A = P_A^0 x_A = 450 	imes 0.4 = 180 \ mm \ Hg$. $P_B = P_B^0 x_B = 700 	imes 0.6 = 420 \ mm \ Hg$. Mole fractions in vapour phase $(y_A, y_B)$: $y_A = P_A / P_{total} = 180 / 600 = 0.3$. $y_B = P_B / P_{total} = 420 / 600 = 0.7$.

At $350 \ K$,the vapour pressure of pure liquids $A$ and $B$ are $450 \ mm \ Hg$ and $700 \ mm \ Hg$ respectively. If the total vapour pressure of the liquid mixture is $600 \ mm \ Hg$,the mole fractions of $A$ and $B$ in the vapour phase respectively are:

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