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,

$P_T=P_{pentane}+P_{hexane}\\\\P_T=X_{pentane}\times P^o_{pentane}+X_{hexane}\times P^o_{hexane}\\\\P_T=(1-X_{hexane})\times P^o_{pentane}+X_{hexane}\times P^o_{hexane}$

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:

$P_T=(1-X_{hexane})\times P^o_{pentane}+X_{hexane}\times P^o_{hexane}$

$270=(1-X_{hexane})\times 425+X_{hexane}\times 151$

$X_{hexane}=0.566$

Thus, the mole fraction of hexane is, 0.566