Benzene and toluene form ideal solution over the entire range of composition. The vapour pressure of pure benzene and toluene at $300 \mathrm{~K}$ are $50.71 \mathrm{~mm} \mathrm{Hg}$ and $32.06$ $\mathrm{mm} \mathbf{H g}$ respectively. Calculate the mole fraction of benzene in vapour phase if $80 \mathrm{~g}$ of benzene is mixed with $100 \mathrm{~g}$ of toluene.

Question

Benzene and toluene form ideal solution over the entire range of composition. The vapour pressure of pure benzene and toluene at $300 \mathrm{~K}$ are $50.71 \mathrm{~mm} \mathrm{Hg}$ and $32.06$ $\mathrm{mm} \mathbf{H g}$ respectively. Calculate the mole fraction of benzene in vapour phase if $80 \mathrm{~g}$ of benzene is mixed with $100 \mathrm{~g}$ of toluene.,

MARKS App · Accepted Answer

Molar mass of $\mathrm{C}_6 \mathrm{H}_6=78 \mathrm{~g} \mathrm{~mol}^{-1}$ Molar mass of $\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_3=92 \mathrm{gmol}^{-1}$ No. of moles of $\mathrm{C}_6 \mathrm{H}_6=\frac{80}{78}=1.026$ mole No. of moles of $\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_3=\frac{100}{92}=1.087 \mathrm{~mole}$ Mole fraction of $\mathrm{C}_6 \mathrm{H}_6$, $x_B=\frac{1.026}{1.026+1.087}=0.486$ Mole fraction of $\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_3$, $x_T=\frac{1.087}{1.026+1.087}=0.514$ According to Raoult's Law, $\begin{aligned} &P_B=x_B \times P_B^{\circ}=0.486 \times 50.71=24.65 \mathrm{~mm} \mathrm{Hg} \ &P_T=x_T \times P_T^{\circ}=0.514 \times 32.06=16.48 \mathrm{~mm} \mathrm{Hg} \end{aligned}$ Mole fraction of $\mathrm{C}_6 \mathrm{H}_6$ in vapour phase $=\frac{P_B}{P_B+P_T}=\frac{24.65}{24 \cdot 65+16 \cdot 48}=0.599 .$

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