Given that the vapor pressure of pure n-hexane and pure n-heptane at 25°c are 151.4 mmhg and 45.62 mmhg respectively, calculate the total vapor pressure above a solution containing only n-hexane and n-heptane in which the mole fraction of n-hexane is 0.600.

Answer is: total pressure is 108.96 mmHg.

p(n-hexane) = 151.4 mmHg; pressure of n-hexane.
p(n-heptane) = 45.62 mmHg.
χ(n-hexane) = 0.600; mole fraction.
χ(n-heptane) = 1 - 0.6 = 0.400.
Using Raoult's Law: 
p(total) = p(n-hexane) · χ(n-hexane)  + p(n-heptane) · χ(n-heptane).
p(total) = 151.4 mmHg · 0.6 + 45.32 mmHg · 0.4.
p(total) = 108.96 mmHg.

What does that mean you dum bum