Question

If the relative lowering of pressure of o-xylene is 0.005 due to addition of 0.5 grams of non-volatile solute in 500 grams of solvent, what is the molecular weight of the solute?

(a) 21.3 g/mole

(b) 23.1 g/mole

(c) 32.1 g/mole

(d) 1.23 g/mole

I have been asked this question at a job interview.

I need to ask this question from Colligative Properties and Determination of Molar Mass in section Solutions of Chemistry – Class 12

Answer 1

The correct option is (a) 21.3 g/mole

Best explanation: Given,

∆p1/p^01 = 0.005

Mass of solute, mW = 0.5 grams

Mass of solvent, mS = 500 grams

Number of moles of solute, nS = 0.5/MW, where MW is the molecular weight of the solute.

Number of moles of solvent, nSolvent = 500 grams/(106 g/mole) = 4.7 mole

Since mS<<mSolvent , the solution can be considered to be dilute. Hence, equation for relative lowering for vapor pressure becomes:

Δp1/p^01 = n2/(n1 + n2) ≈ n2/n1

On substituting values,

0.005 = (0.5/MW)/4.7

On solving, MW = 21.3 g/mole.