Question

A solution contains a mixture of pentane and hexane at room temperature. The solution has a vapor pressure of 256 {\rm torr}. Pure pentane and hexane have vapor pressures of 425 {\rm torr} and 151 {\rm torr}, respectively, at room temperature.What is the mole fraction of hexane? (Assume ideal behavior.)

Answer 1

Answer: The mole fraction of hexane is 0.617

Step-by-step explanation:

According to Raoult's law, the vapor pressure of a component at a given temperature is equal to the mole fraction of that component multiplied by the vapor pressure of that component in the pure state.

`p_1=x_1p_1^0` and
`p_2=x_2P_2^0`

where, x = mole fraction in solution

`p^0` = pressure in the pure state

According to Dalton's law, the total pressure is the sum of individual pressures.

`p_(total)=p_1+p_2`

`p_(total)=x_(pentane)p_(pentane)^0+x_(hexane)P_(hexane)^0`

`x_(pentane)=x`

`x_(hexane)=1-x_(pentane)=1-x`

`p_(pentane)^0=425torr`

`p_(hexane)^0=151torr`

`256torr=x* 425+(1-x)* 151`

`x=0.383`

Thus mole fraction of hexane is (1-0.383) = 0.617 .