Calculate the vapour pressure of pure volatil | vedclass

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Question

(D) According to Raoult's Law for a binary solution of two volatile liquids $A$ and $B$: $P_{total} = P_A + P_B = P_A^0 \chi_A + P_B^0 \chi_B$ Given:

Vedclass · Accepted Answer

$400$

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(D) According to Raoult's Law for a binary solution of two volatile liquids $A$ and $B$: $P_{total} = P_A + P_B = P_A^0 \chi_A + P_B^0 \chi_B$ Given: $P_B^0 = 900 \ mm \ Hg$ $\chi_B = 0.4$ $P_{total} = 600 \ mm \ Hg$ Since $\chi_A + \chi_B = 1$,we have $\chi_A = 1 - 0.4 = 0.6$. Substituting the values into the equation: $600 = P_A^0(0.6) + (900)(0.4)$ $600 = P_A^0(0.6) + 360$ $P_A^0(0.6) = 600 - 360 = 240$ $P_A^0 = \frac{240}{0.6} = 400 \ mm \ Hg$ Thus,the vapour pressure of pure liquid $A$ is $400 \ mm \ Hg$.

Calculate the vapour pressure of pure volatile liquid $A$ at a given temperature if the mole fraction and vapour pressure of pure volatile liquid $B$ are $0.4$ and $900 \ mm \ Hg$ respectively,given that the total vapour pressure of the solution is $600 \ mm \ Hg$. (in $mm \ Hg$)

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