Question

At $300K$ the vapour pressure of an ideal solution containing $1$ mole of liquid $A$ and $2$ moles of liquid $B$ is $500mm$ of $Hg$. The vapour pressure of the solution increases by $25mm$ of $Hg$, if one more mole of $B$ is added to the above ideal solution at $300K$. Then the vapour pressure of $A$ in its pure state is

• A. $300mm$ of $Hg$
• B. $400mm$ of $Hg$
• C. $500mm$ of $Hg$
• D. $600mm$ of $Hg$

Answer 1

Answer: (A)

According to Raoult's law,
$P_T=x_A{p}_A^{\circ }+x_B{p}_B^{\circ }$
Given, $P_{T_1}=500mmHg$
$n_A=1$ and $n_B=2$
$\therefore x_A=1/3$ and $x_B=2/3$
$\Rightarrow 500=\frac{1}{3}{p}_A^{\circ }+\frac{2}{3}{p}_B^{\circ }$
$\Rightarrow 1500=p_A^{\circ }+2p_B^{\circ }....$(i)
Also given that, one more mole of $B$ is added to the solution, the pressure of the ideal solution increases by $25mmHg$.
$\therefore P_{T_2}=500+25=525mmHg$
Also, $n_B=3\therefore x_A=1/4$ and $x_B=3/4$
$525=\frac{1}{4}{p}_A^{\circ }+\frac{3}{4}{p}_B^{\circ }.....$(ii)
$2100=p_A^{\circ }+3p_B^{\circ }$
Subtract (i) and (ii),
$p_B^{\circ }=600mmHg$
$p_A^{\circ }+2p_B^{\circ }=1500$
$\Rightarrow p_A^{\circ }=300mmHg$