Question

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

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

Answer 1

Answer: (A)

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