Question

The vapor pressure of benzene, c6h6, is 100.0 torr at 26.1 °c. Assuming raoult's law is obeyed, how many moles of a nonvolatile solute must be added to 100.0 ml of benzene to decrease its vapor pressure by 10.0% at 26.1 °c? The density of benzene is 0.8765 g>cm3.

Answer 1

According to Raoult's law:

`p_(solution) = X_(solvent)* P_(solvent)` -(1)

where
`p_(solution)` is observed vapor pressure of the solution,
`X_(solvent)` is mole fraction of solvent, and
`P_(solvent)` is vapor pressure of the pure solvent.

Vapor pressure of benzene =
`100 torr at 26.1^(o) C` (given)

Reduction in vapor pressure on addition of non-volatile solute = 10.0 % (given)

So, vapor pressure of the component in the solution =
`100 - 10 = 90 torr`

Substituting the values in formula (1):

`90 = mole fraction of benzene * 100`

`mole fraction of benzene = (90)/(100) = 0.9`

Mole of benzene =
`Volume* (density)/(Molar Mass)`

Mole of benzene =
`100 cm^(3)* (0.8765 g/cm^(3))/(78 g/mol) = 1.124 mol`

Since,
`mole fraction of benzene =(mole of benzene )/(mole of benzene +mole of non volatile solute )`

So,
`0.9 =(1.124 )/(1.124+mole of non volatile solute )`

`(1.124)/(0.9) =1.124+mole of non volatile solute`

`mole of non volatile solute = 1.249 - 1.124 = 0.125`

Hence, moles of a nonvolatile solute to be added is
`0.125`.