Question

An ideal mixture of liquids $A$ and $B$ with 2 moles of $A$ and 2 moles of $B$ has a total vapour pressure of $1atm$ at a certain temperature. Another mixture with 1 mole of $A$ and 3 moles of $B$ has vapour pressure greater than 1 atm. When 4 moles of $C$ are added to second mixture, the vapour pressure comes down to 1 atm. Vapour pressure of $C$ in pure state i.e., $P_{\circ }^{\circ }=0.8atm.$ What will be the value of $P_A^{\circ }\&P_B^{\circ }$

• A. $P_A^{\circ }=1.2atm,P_B^o=0.7atm$
• B. $P_A^{\circ }=1.2atm,P_B^{\circ }=0.6atm$
• C. $P_A^{\circ }=0.6atm,P_B^{\circ }=1.4atm$
• D. $P_A^{\circ }=1.4atm,P_B^{\circ }=0.6atm$

Answer 1

Answer: (C)

$1=P_A+P_B$

$1=P_A^{\circ }\cdot \frac{2}{4}+P_B^{\circ }\cdot \frac{2}{4}$

$P_A^{\circ }+P_B^{\circ }=2\dots \dots$(1)

$x=P_A^{\circ }{x}_A+P_B^{\circ }{x}_B$

$x=P_A^{\circ }\frac{1}{4}+P_B^{\circ }\frac{3}{4}$

$P_A^{\circ }+3P_B^{\circ }=4x\dots$(2)

$1=P_A^{\circ }{x}_A+P_B^{\circ }{x}_B+P_c^{\circ }{x}_C$

$1=P_A^{\circ }\cdot \frac{1}{8}+P_B^{\circ }\cdot \frac{3}{8}+P_C^{\circ }\cdot \frac{4}{8}$

$8=P_A^{\circ }+3P_B^{\circ }+3.2$

$P_A^{\circ }+3P_B^{\circ }=4.8\dots \dots (3)$

From 2 and $3:4x=4.8$

$x=1.2$

Also, $1-2=P_A^{\circ }+P_B^{\circ }-\left(P_A^{\circ }+3P_B^{\circ }\right)=2-4x$

$=P_A^{\circ }+P_B^{\circ }-P_A^{\circ }-3P_B^{\circ }=2-4x$

$=-2P_B^{\circ }=2-4x$

$P_B^{\circ }=4x-2$

$P_B^{\circ }=4\times 1.2-2$

$P_B^{\circ }=4.8-2$

$P_B^{\circ }=\frac{2.8}{2}=1.4$

$P_A^{\circ }+P_B^{\circ }=2$

$\Rightarrow P_A^{\circ }+1.4=2$

$P_A^{\circ }=0.6$