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

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.

$p_{total}=p_1+p_2$

$p_{total}=x_{pentane}p_{pentane}^0+x_{hexane}P_{hexane}^0$

$x_{pentane}=x$ 

$x_{hexane}=1-x_{pentane}=1-x$ 

$p_{pentane}^0=425torr$

$p_{hexane}^0=151torr$

$256torr=x\times 425+(1-x)\times 151$

$x=0.383$

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