Question

The vapour pressure of pure benzene at a certain temperature is 640 mm Hg. A non-volatile solute weighing 2.175 g is added to 39.0 g of benzene. The vapour pressure of the solution is 600 mm Hg. What is the molar mass of the solute?

Answer 1

Given: Vapour pressure of pure benzene (P°) = 640 mm Hg

Vapour pressure of solution (P) = 600 mm Hg

Mass of solute (w₂) = 2.175 g

Mass of solvent (benzene) (w₁) = 39.0 g

Molar mass of benzene (M₁) = 78 g/mol

Molar mass of solute = M₂​ (to be calculated)

By using the formula for the Relative lowering of vapour pressure

`(P^circ - P)/P^circ = (w_2 * M_1)/(w_1 * M_2)`

⇒ `(640 - 600)/640 = (2.175 xx 78)/(39 xx M_2)`

⇒ `40/640 = 169.65/(39 xx M_2)`

⇒ 0.0625 = `169.65/(39 xx M_2)`

⇒ `0.0625 xx 39 = 169.65/M_2`

⇒ `2.4375 = 169.65/M_2`

⇒ `M_2 = 169.65/2.4375`

M₂ = 69.6 g/mol

∴ The molar mass of the solute is 69.6 g/mol.