Question

Equal moles of benzene and toluene are mixed the V.P. of benzene and toluene in pure state are 700 and 600 mm Hg respectively. The mole fraction of benzene in vapour state is:

• A. 0.7
• B. 0.47
• C. 0.5
• D. 0.54

Answer 1

The correct option is D 0.54

Suppose the moles of Benzene & Toluene are $x$.

Hence, their mole fraction$=\frac{x}{2x}=\frac{1}{2}$

Vapour pressure of Benzene in pure state $P\left(a\right)=700mmofHg$

and that of Toluene $P\left(b\right)=600mmofHg$

Now total pressure$P\left(t\right)=P\left(a\right)\times \left(a\right)\times P\left(b\right)\times \left(b\right)$

$=700\times \frac{1}{2}+600\times \frac{1}{2}=650mmofHg$

Now, $P\left(a\right)\times \left(a\right)=P\left(t\right)Y\left(a\right)$ where $Y\left(a\right)=molefractionof{C}_6{H}_{6}invapourstate$

$\Rightarrow 700\left(\frac{1}{2}\right)=650\left(Y\right(a\left)\right)$

$\therefore Y\left(a\right)=\frac{350}{650}=0.538\approx 0.54$

Hence, the correct option is $D$