Question text What is the mole fraction of benzene in a benzene-toluene solution whose vapor pressure is 51 torr at 20°C?

Mole fraction for benzene is 0.683

Explanation:

We need to determine the colligative property about vapour pressure which formula is:

ΔP = P° . Xm

where P° is vapour pressure of pure solvent.

ΔP = Vapor pressure of pure solvent - Vapor pressure of solution

In this case, our solvent is benzene and the solute is toluene. For this colligative property vapour pressure of solution is lower than pure solvent.

P° of benzene at 20°C is 74.7 Torr

Xm is the value for the mole fraction of solute.

Let's replace data: 74.7 Torr - 51 Torr = 74.7 Torr . Xm

Xm = 0.317 → That's the mole fraction for toluene.

Then, mole fraction for benzene will be: 1 - 0.317 → 0.683

when the volume of gas is   changed from   717.73 ml   to 852 ml   the temperature will   change   from 315   c to 452   c

    calculation

this is calculated using the charles law formula

that is,   v1/t1 = v2/t2

          v1=?

        t1 = 315 c   into   kelvin = 315 +273 = 588 k

        v2 = 852   ml

        t2 = 425c + 273 =698 k

make v1   the subject of the formula ( by   multiplying both side by t1)

  v1=( v2 xt1) / t2

v1= (852 ml   x 588 k)/ 698 k= 717.73 ml