Question

Vapour pressure of benzene is 200 mm of Hg. When 2 gram of a
non-volatile solute dissolved in 78 gram benzene. Benzene has
vapour pressure of 195 mm of Hg. Calculate the molar mass of the
solute. (molar mass of benzene is 78 gram mor']​

Answer 1

The Molar mass of solute is : 12.974 g/mol.

Given, Vapour pressure of pure Benzene = 200 mm of Hg.

            Mass of Solute = 2 gm, Mass of Solvent = 78 gm,                                           Vapour Pressure = 195 mm of Hg, Molar mass of Benzene = 78 gm/mol, kb for benzene = 2.53 Kkg/mol.

We know, Depression in freezing point

Δ$T_{b}$  = Pure solvent vapour pressure - Vapour pressure of soute added solvent.

Δ$T_{b}$ = 200 - 195 = 5 mm of Hg.

Also,

Δ$T_{b}$ = kb * weight of solute *1000 / Molar mass of solute * mass of solvent

   5  = 2.53 * 2 * 1000 / M * 78

    5 = 64.8717 / M,    M = 64.8717/5 = 12.974

Therefore Molar mass of solute = 12.974 g/mol.

Answer 2

The molar mass of the  solute is 80 g/mol

Given:

p⁰ = 200 mm Hg

p = 195 mm Hg

Weight of solute = w = 2 g

Weight of solvent = w₀ = 78 g

Molar mass of benzene = M = 78 g/mol

To find:

Molecular weight of solute = m = ?

Explanation:

The vapor pressure is given by the relationship given below:

(p⁰ - p)/p⁰ = (w × M)/(w₀ × m)

On substituting the values, we get,

(200 - 195)/200 = (2 × 78)/(78 × m)

5/200 = 156/78m

m = 156/78 × 200/5

m = 2 × 40

∴ m = 80 g/mol