Solutions Question 171
Question: Vapour pressure of pure benzene is 119 torr and that of toluene is 37.0 torr at the same temperature. Mole fraction of toluene in vapour phase which is in equilibrium with a solution of benzene and toluene having a mole fraction of toluene 0.50, will be:
Options:
A) 0.137
B) 0.237
C) 0.435
D) 0.205
Show Answer
Answer:
Correct Answer: B
Solution:
$ P_{A}=P_A^{{}^\circ }\times x_{A} $ = total pressure $ \times y_{A} $
$ P_{B}=P_B^{{}^\circ }\times x_{B} $ pressure $ \times y_{B} $ where x and y represents mole fraction in liquid and vapour phase respectively. $ \frac{P_B^{{}^\circ }x_{B}}{P_A^{{}^\circ }x_{A}}=\frac{y_{B}}{y_{A}};\frac{P_B^{{}^\circ }(1-x_{A})}{P_A^{{}^\circ }x_{A}}=\frac{1-y_{A}}{y_{A}} $ on putting values $ \frac{119(1-0.50)}{37\times 0.50}=\frac{1-y_{A}}{y_{A}} $ on solving $ y_{A}=0.237 $
BETA
