Question

Two volatile liquids $A$ and $B$ an mixed in mole ratio $1:3$, to form an ideal liquid solution. Vapour pressure of pure $A$ and $B$ are $600torr$ and $200torr$. Volume of container for vapour is $76$ litre. $(R=1/12$ litre atm/K/mole).

Calculate the mole of $\wedge$ in vapour phase at $300K$, assuming negligible amount of vapour is present compared to liquid in container.

• A. $2mole$
• B. $0.3mole$
• C. $1.2mole$
• D. $0.6mole$

Answer 1

Answer: (B)

$0.3mole$

$A$ and $B$ are mixed in mole ratio $1:3$

Total vapour pressure of the solution is,

$P=p_A+p_B=x_A{p}_A^{\ast }+x_B{p}_B^{\ast }$

$=\frac{1}{4}\times 600+\frac{3}{4}\times 200$ torr

$=300$ torr $=0.3947$ atom

now $PV=nRT$

$\Rightarrow n=\frac{PV}{RT}=\frac{0.3947\times 76}{1/12\times 300}=1.2$

$\therefore$ total moles, $n_A+n_B=1.2$

$\therefore n_A=\frac{1}{4}\times 1.2=0.3$ mole