When air was passed through glass bulb containing 16 g of an organic compound in 100 g of water and then through glass bulb containing only water at the same temperature and then finally through anhydrous CCl4 glass tube, the loss in mass of glass bulb containing water was 0.052 g and the gain in mass of CCl4 tube was 1.86 g. Calculate the molecular mass of the organic substance.
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 100 g
This is the gas-saturation (Ostwald-Walker) method of measuring the relative lowering of vapour pressure. Dry air picks up vapour in proportion to the vapour pressure it meets:
In the solution bulb it takes up vapour ∝ ps (vapour pressure of the solution).
In the water bulb it takes up the extra amount ∝ (p° − ps), so the loss in mass of the water bulb ∝ p° − ps.
All the water vapour is absorbed in the drying tube (anhydrous \( \ce{CaCl2} \)), so the gain in mass of that tube ∝ p°.
\( \displaystyle \frac{p^{\circ}-p_s}{p^{\circ}} = \frac{\text{loss in water bulb}}{\text{gain in drying tube}} = \frac{0.052}{1.86} = 0.02796 \)
By Raoult’s law the relative lowering equals the mole fraction of the solute:
\( \displaystyle \frac{n_2}{n_1+n_2} = 0.02796,\qquad n_1 = \frac{100}{18} = 5.556\ \mathrm{mol} \)
\( \displaystyle n_2 = \frac{0.02796\times5.556}{1-0.02796} = 0.1598\ \mathrm{mol} \)
\( \displaystyle M_2 = \frac{16\ \mathrm{g}}{0.1598\ \mathrm{mol}} \approx 100\ \mathrm{g\,mol^{-1}} \)
So the molecular mass of the substance is about 100 (printed as 100 g).
Note 1: The question is from beyond the NCERT textbook.
Note 2: It seems that there is a mistake in the original NTA paper, which prints anhydrous CCl4 when it should read anhydrous CaCl2.