Question

What is the vapor pressure of isopropanol (MW = 60.1) in a closed container of 70% isopropanol in water (MW = 18.0) at 25oC? Such a mixture contains 70 g of isopropanol and 30 g of water. The vapor pressure of pure isopropanol at 25oC = 50 torr

Answer 1

The vapor pressure of the 70% isopropanol solution is 21 Torr.

Step-by-step explanation:

Relative lowering of vapor pressure for solution with non volatile solute is given by :

`(p^o-p)/(p^o)=\chi_(solute)`

Where:

`p^o` = Vapor pressure of pure solvent

p = Vapor pressure of solution

`\chi_(solute)` = Mole fraction of solute

We have, 70% isopropanol solution which menas 70 grams of isopropanol (solvent) and 30 grams of water (solute).

Mass of water = 30 g

Moles of water =
`n_w=(30 g)/(18.0 g/mol)=1.667 mol`

Mass of isopropanol = 70 g

Moles of isopropanol =
`n_i=(70 g)/(60.1 g/mol)=1.165 mol`

Mole fraction of solute that is water :

`\chi_w=(1.667 mol)/(1.667 mol+1.165 mol)=0.5886`

Vapor pressure of the pure isopropanol =
`p^o=50 Torr`

Vapor pressure of the solution = p

`(50 Torr-p)/(50 Torr)=0.5886`

Solving for p:

p = 20.57 Torr ≈ 21 Torr

The vapor pressure of the 70% isopropanol solution is 21 Torr.