At \(300 \mathrm{~K}\) an ideal solution is formed by mixing \(460 \mathrm{~g}\) of toluene with \(390…
At \(300 \mathrm{~K}\) an ideal solution is formed by mixing \(460 \mathrm{~g}\) of toluene with \(390 \mathrm{~g}\) benzene. If the vapour pressure of pure toluene and benzene at \(300 \mathrm{~K}\) are 32 and \(40 \mathrm{~mm}\) respectively, the mole fraction of toluene in vapour phase is
0.196
0.588
0.294
0.444
Solution
Given, \(w_a=460 \mathrm{~g}\) (toluene)
\(w_B=390 \mathrm{~g} \text { (benzene) }\)
Vapour pressure of pure toluene \(\left(p_A\right)=32 \mathrm{~mm}\)
Vapour pressure of pure benzene \(\left(p_B\right)=40 \mathrm{~mm}\)
Moles of toluene, \(\left(n_A\right)=\frac{w_A}{M_A}=\frac{460}{92}\)
Moles of benzene, \(n_B=\frac{w_B}{M_B}=\frac{390}{78}\)
Hence, mole fraction of toluene \(\chi_A\).
\(\chi_A=\frac{\frac{460}{92}}{\frac{460}{92}+\frac{390}{78}}\)
Mole fraction of benzene \(\left(\chi_B\right)\)
\(\begin{gathered}
\chi_B=\frac{\frac{390}{78}}{\frac{460}{92}+\frac{390}{78}} \\
\therefore \quad p_{\text {total }}=p^{\circ}{ }_A \chi_A+p^{\circ}{ }_B \chi_B \\
p_{\text {total }}=32 \times\left(\frac{\frac{460}{92}}{\frac{460}{92}+\frac{390}{78}}\right)+40\left(\frac{\frac{390}{78}}{\frac{390}{78}+\frac{460}{92}}\right) \\
p_{\text {total }}=36 \mathrm{~mm}
\end{gathered}\)
Mole fraction of toluene in vapour phase \(\left(Y_T\right)\) can be calculated as below :
\(p_T^{\circ} \chi_T=Y_T, p_{\text {total }} \Rightarrow Y_T=\frac{p_T^{\circ} \chi_T}{p_{\text {total }}}\)
where, \(\chi_T\) is mole fraction of toluene.
\(Y_T=\frac{32 \times \frac{1}{2}}{36} \Rightarrow Y_T=\frac{16}{36}=0.444\)