Question

Two liquids $A$ and $B$ form ideal solutions. At $300K$, the vapour pressure of solution containing $1$ mole of $A$ and $3$ mole of $B$ is $550mmHg$. At the same temperature, if one more mole of $B$ is added to this solution, the vapour pressure of the solution increases by $10mmHg$. Determine the vapour pressure of $A$ and $B$ in the pure states (in $mmHg$).

• A. $400,600$
• B. $500,500$
• C. $600,400$
• D. None of these

Answer 1

Answer: (A)

$400,600$

According to Raoult's law,
$P_{total}=P_A^0\times x_A+P_B^0\times x_B$

1 mole of A and 3 mole of B corresponds to the mole fractions 0.25 and 0.75 respectively.

Substitute values in the above expression.
$550=P_A^0\times 0.25+P_B^0\times 0.75$ ......(1)

When one mole of B is added, the mole fractions become 0.2 and 0.8 respectively.
$550+10=P_A^0\times 0.20+P_B^0\times 0.8$ ......(2)

Equations (1) and (2) are solved to obtain then vapour pressures of A and B in the pure state.

$P_A^0=400mmHg$
$P_B^0=600mmHg$