At 20 ? c the vapor pressure of benzene (c6h6)is 75 torr, and that of toluene (c7h8) is 22 torr

. assume that benzene and toluene form an ideal solution.

what is the composition in mole fractions of a solution that has a vapor pressure of 36torr at 20 degrees c?

what is the mole fraction of benzene in the vapor above the solution described in part (a)?

(a) Benzene = 0.26; toluene = 0.74

(b) Benzene = 0.55

Explanation:

1. Calculate the composition of the solution

For convenience, let’s call benzene Component 1 and toluene Component 2.

According to Raoult’s Law,  

where

p₁ and p₂ are the vapour pressures of the components above the solution

χ₁ and χ₂ are the mole fractions of the components

p₁° and p₂° are the vapour pressures of the pure components.

Note that

χ₁ + χ₂ = 1

So,  

χ₁ = 0.26 and χ₂ = 0.74

2. Calculate the mole fraction of benzene in the vapour

In the liquid,  

p₁ = χ₁p₁° = 0.26 × 75 mm = 20 mm

∴ In the vapour

Note that the vapour composition diagram below has toluene along the horizontal axis. The purple line is the vapour pressure curve for the vapour. Since χ₂ has dropped to 0.45, χ₁ has increased to 0.55.

a: 45.9 meters

b: 4.29 seconds

step-by-step explanation:

a: h(t)=1/2(-9.8)(3)^2+0(3)+90

h(t)=-4.9(3)^2+90

h(t)=-44.1+90

h(t)=45.9 meters

b: 0=1/2(-9.8)(t^2)+0t+90

0=-4.9t^2+90

-90=-4.9t^2

18.37=t^2

t=4.29 seconds

in the second question, the numbers aren't exact, but they're a close approximation.

c, liquids evaporate, creating a gas