Question

At $300K$, the vapour pressure of an ideal solution that has 3 mol A and 2 mol B is $600torr$. At the same temperature, if $1.5$ mol A & $0.5$ mol C (non-volatile) are added to this solution, the vapour pressure of the solution increases by 30 torr. Then, what is the value of $P_B^{\circ }$?

• A. $940torr$
• B. $405torr$
• C. $90torr$
• D. $45torr$

Answer 1

The correct option is C $90torr$

According to Raoult's law$P=P_A^{\circ }{X}_A+P_B^{\circ }{X}_{B}600=P_A^{\circ }\left(\frac{3}{3+2}\right)+P_B^{\circ }\left(\frac{2}{2+3}\right)$

$3000=3P_A^{\circ }+2P_B^{\circ }$ ......(1)

After the addition of 1.5 moles of A and 0.5 mole of C,

$630=P_A^{\circ }\left(\frac{4.5}{4.5+2+0.5}\right)+P_B^{\circ }\left(\frac{2}{4.5+2+0.5}\right)$

$4.5P_A^{\circ }+2P_B^{\circ }=4410$ .......(2)

Solving equation (1) and (2) we get,
$1.5P_A^{\circ }=1410⟹P_A^{\circ }=940torr$ & $P_B^{\circ }=90torr$