A) 0.467
B) 0.502
C) 0.513
D) 0.556
Correct Answer: C
Solution :
\[{{P}_{m}}=P_{MeOH}^{0}X_{MeOH}^{0}+P_{EtOH}^{0}X_{EtOH}^{{}}\] \[{{P}_{m}}=88.5\times \left[ \frac{\frac{16}{32}}{\frac{16}{32}+\frac{46}{46}} \right]+42X;\,\,\left[ \frac{\frac{46}{46}}{\frac{16}{32}+\frac{46}{46}} \right]\,\,=\,\,57.5\]Now \[P{{'}_{MeOH}}=={{P}_{m}}\cdot {{X}_{MeOH(g)}}\] \[\therefore \,\,\,\,\,88.5\,\times =57.5\,\,{{X}_{MeOH(g)}}\] \[K=\left[ \frac{\frac{16}{32}}{\frac{16}{32}+\frac{46}{46}} \right]\] \[\therefore \,\,\,\,\,\,\,\,{{X}_{MeOH}}\,\,=\,\,0.513\]You need to login to perform this action.
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