Question

A solution contains a mixture of pentane and hexane at room temperature. The solution has a vapor pressure of 270 torr . Pure pentane and hexane have vapor pressures of 425 torr and 151 torr, respectively, at room temperature.

Answer 1

This is an incomplete question, here is a complete question.

A solution contains a mixture of pentane and hexane at room temperature. The solution has a vapor pressure of 270 torr . Pure pentane and hexane have vapor pressures of 425 torr and 151 torr, respectively, at room temperature.

What is the mole fraction of hexane?

Answer : The mole fraction of hexane is, 0.566

Explanation :

According to Raoult's law,

`P_T=P_(pentane)+P_(hexane)\\\\P_T=X_(pentane)* P^o_(pentane)+X_(hexane)* P^o_(hexane)\\\\P_T=(1-X_(hexane))* P^o_(pentane)+X_(hexane)* P^o_(hexane)`

where,

`P_T` = total vapor pressure = 270 torr

`P^o_(pentane)` = vapor pressure of pure pentane = 425 torr

`P^o_(hexane)` = vapor pressure of pure hexane= 151 torr

`X_(hexane)` = mole fraction of hexane = ?

`X_(pentane)` = mole fraction of pentane

Now put all the given values in the above formula, we get:

`P_T=(1-X_(hexane))* P^o_(pentane)+X_(hexane)* P^o_(hexane)`

`270=(1-X_(hexane))* 425+X_(hexane)* 151`

`X_(hexane)=0.566`

Thus, the mole fraction of hexane is, 0.566