The ideal solutions formed by mixing two liqu | vedclass

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(C) For an ideal solution,the total vapour pressure $P_T$ is given by Raoult's law: $P_T = P_A^0 x_A + P_B^0 x_B$,where $x_A + x_B = 1$.Case $1$: Mola

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$550, 250$

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(C) For an ideal solution,the total vapour pressure $P_T$ is given by Raoult's law: $P_T = P_A^0 x_A + P_B^0 x_B$,where $x_A + x_B = 1$.Case $1$: Molar ratio $1:1$,so $x_A = 0.5$ and $x_B = 0.5$. $P_T = 400 \ mm$. Thus,$0.5 P_A^0 + 0.5 P_B^0 = 400$,which simplifies to $P_A^0 + P_B^0 = 800$ (Equation $1$).Case $2$: Molar ratio $1:2$,so $x_A = 1/3$ and $x_B = 2/3$. $P_T = 350 \ mm$. Thus,$(1/3) P_A^0 + (2/3) P_B^0 = 350$,which simplifies to $P_A^0 + 2 P_B^0 = 1050$ (Equation $2$).Subtracting Equation $1$ from Equation $2$: $(P_A^0 + 2 P_B^0) - (P_A^0 + P_B^0) = 1050 - 800$,so $P_B^0 = 250 \ mm$.Substituting $P_B^0 = 250$ into Equation $1$: $P_A^0 + 250 = 800$,so $P_A^0 = 550 \ mm$.Therefore,the vapour pressures of pure liquids $A$ and $B$ are $550 \ mm$ and $250 \ mm$ respectively.

The ideal solutions formed by mixing two liquids $A$ and $B$ at $300 \ K$ in the molar ratio of $1:1$ and $1:2$ have vapour pressures of $400 \ mm$ and $350 \ mm$ respectively. At the same temperature,the vapour pressures of pure liquids $A$ and $B$ in $mm$ respectively are:

Similar Questions

What is meant by positive and negative deviations from Raoult's law and how is the sign of $\Delta_{mix} H$ related to positive and negative deviations from Raoult's law?

Two compounds form an ideal solution at room temperature. Which of the following are correct for this ideal solution?
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An ideal solution has two components $A$ and $B$. If $A$ is more volatile than $B$,then the correct relation between mole fraction of $A$ in liquid $(X_A)$ and vapour phase $(Y_A)$ is:

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