A) 0.549
B) 0.200
C) 0.786
D) 0.478
Correct Answer: D
Solution :
| [d] Total vapour pressure of mixture |
| = (Mole fraction of pentane V.P. of pentane) + (Mole fraction of hexane x V.P. of hexane) = V.P. of pentane in mixture + V.P. of hexane in mixture |
| \[=\left( \frac{1}{5}\times 440+\frac{4}{5}\times 120 \right)=184\,mm\] |
| \[\because \,\] V.P. of pentane in mixture |
| = V.P. of mixture x mole fraction of pentane it vapour phase |
| 88 = 184 x mole fraction of pentane in vapour phase |
| \[\therefore \] Mole fraction of pentane in vapour phase |
| \[=\frac{88}{184}=0.478\] |
You need to login to perform this action.
You will be redirected in
3 sec
