Question

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

• A. $150$
• B. $405$
• C. $90$
• D. none of these

Answer 1

The correct option is A $150$

Mole fraction of B is $X_B=\frac{2}{2+3}=0.4$

The mole fraction of A is $X_A=1-0.4=0.6$

The expression for the total pressure of the solution is

$P=P_A^o{X}_A+P_B^o{X}_B$

Substitute values in the above expression

$P=600$ torr $=P_A^o\times 0.6+P_B^o\times 0.4$ ......(1)

1.5 mole of A & 0.5 mole of B (non-volatile) are added

Mole fraction of B is $X_B=\frac{2.5}{5+2}=0.36$

The mole fraction of A is $X_A=1-0.36=0.64$

The expression for the total pressure of the solution is

$P=P_A^o{X}_A+P_B^o{X}_B$

The total pressure is $600+30=630torr$

Substitute values in the above expression

$P=630torr=P_A^o\times 0.64+P_B^o\times 0.36$ ......(2)

Rearrange equation (2)

$P_A^o=\frac{630-0.36P_B^o}{0.64}$.....(3)

From (1) and (3),

$600=0.6\left(\frac{630-0.36P_B^o}{0.64}\right)+0.4P_B^o$

$600=590.625-0.3375P_B^o+0.4P_B^o$

$600=590.625+0.0625P_B^o$

$P_B^o=\frac{600-590.625}{0.0625}=150torr$

Hence, the correct option is $\text{A}$