It is desired to prepare a mixture of n-hexane: (T_bp = 68.74*C) and n-octane (T

Question

It is desired to prepare a mixture of n-hexane: (T_bp = 68.74*C) and n-octane (T_bp = 125.5*C) at standard pressure and 90 degree C in a sealed container. Under these conditions, both vapor and liquid are present. At 90 degree C the vapor pressure of n-hexane is 1417 mmHg, and the vapor pressure for n-octane is 250.8 mmHg. What is the composition of the liquid? What is the composition of the vapor? If there is 0.5 mol of n-hexane and 0.5 of n-octane total in the container, how many moles exist in the liquid phase?

Explanation / Answer

Apply Raoutls law

x1*P°1 = y1*Pt

x2*P°2 = y2*Pt

and y1+y2 = 1 & x1 + x2 = 1

so:

x1*P°1 = y1*Pt

(1-x1)*P°2 = (1-y1)*Pt

substitute known data

x1*1417 = y1*760

(1-x1)*250.8 = (1-y1)*760

solve for x1:

x1 = 760/1417*y1

x1 = 0.536*y1

substitute

(1-0.536*y1) = (1-y1)*760/250.8

1-0.536y1 = 3.03 - 3.03y1

y1(3.03-0.536) = 3.03-1

y1 = (3.03-1)/(3.03-0.536) = 0.81395

so

y2 = 1-0.81395 = 0.18605

x1 = 0.536*y1 = 0.536*0.81395 = 0.436277

x2 = 1-x1 = 1-0.436277 = 0.563723

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