A mixture of 60 mol % n-propylcyclohexane and 40 mol % n-propylbenzene is distil

Question

A mixture of 60 mol % n-propylcyclohexane and 40 mol % n-propylbenzene is distilled through a simple simple distillation apparatus; assume that no fractionation occurs during the distillation. The boiling temperature is found to be 157 C(760 torr) as the first small amount of distillate is collected. The standard vapor pressures of n-propylcyclohexane and n-propylbenezene are known to be 769 toor and 725 torr, respectively, at 157.3 c. Calculate the percentage of each of the two components in the first few drops of distillate.

Explanation / Answer

Answer:

Vapour pressure of n-propylcyclohexane = 769 torr

Vapour pressure of n-propylbenzene = 725 torr

Ptotal = 769+725

=1494 torr

=751 torr

Let x is the percentage of n-propylcyclohexane, then (1-x) will be the percentage of n-propylbenzene.

Ptotal =[(769*x)+725*(1-x) ]

1494=769x+725-725x

1494-725 = 44x

x = 769/44

Percentage of n-propylcyclohexane =17.477%

Percentage of n-propylbenzene = 100- 17.477 = 82.523%

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