At $80^{\circ} \mathrm{C}$, the vapour pressure of pure liquid ' $A$ ' is $520 \mathrm{~mm} \mathrm{Hg}$ and…

At $80^{\circ} \mathrm{C}$, the vapour pressure of pure liquid ' $A$ ' is $520 \mathrm{~mm} \mathrm{Hg}$ and that of pure liquid ' $B$ ' is $1000 \mathrm{~mm}$ $\mathrm{Hg}$. If a mixture solution of ' $A$ ' and ' $B$ ' boils at $80^{\circ} \mathrm{C}$ and $1 \mathrm{~atm}$ pressure, the amount of ' $A$ ' in the mixture is $(1 \mathrm{~atm}=760 \mathrm{~mm} \mathrm{Hg})$
  1. 52 mol percent
  2. 34 mol percent
  3. 48 mol percent
  4. 50 mol percent

Solution

$ \begin{aligned} & P_T=P_A^{\circ} X_A+P_B^{\circ} X_B \\ & 760=520 X_A+P_B^{\circ}\left(1-X_A\right) \\ & \Rightarrow X_A=0.5 \end{aligned} $ Thus, mole $\%$ of $A=50 \%$

Asked in: JEE Main 2008

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