Respuesta :
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,
[tex]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}[/tex]
where,
[tex]P_T[/tex] = total vapor pressure = 270 torr
[tex]P^o_{pentane}[/tex] = vapor pressure of pure pentane = 425 torr
[tex]P^o_{hexane}[/tex] = vapor pressure of pure hexane= 151 torr
[tex]X_{hexane}[/tex] = mole fraction of hexane = ?
[tex]X_{pentane}[/tex] = mole fraction of pentane
Now put all the given values in the above formula, we get:
[tex]P_T=(1-X_{hexane})\times P^o_{pentane}+X_{hexane}\times P^o_{hexane}[/tex]
[tex]270=(1-X_{hexane})\times 425+X_{hexane}\times 151[/tex]
[tex]X_{hexane}=0.566[/tex]
Thus, the mole fraction of hexane is, 0.566
