20 g of a non-volatile and non-electrolytic solute dissolved in 100 g of water produces the relative lowering of vapor pressure of 0.1. The molar mass of solute in g mol-1 is:
Assume n2 << n1
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 36
When a non-volatile solute is dissolved in a solvent, the vapour pressure falls. By Raoult’s law, the relative lowering of vapour pressure equals the mole fraction of the solute:
\( \displaystyle \frac{p^{0}-p}{p^{0}}=x_2=\frac{n_2}{n_1+n_2} \)
For a dilute solution n2 << n1, so n2 can be neglected in the denominator. Writing n = w/M:
\( \displaystyle \frac{p^{0}-p}{p^{0}}\approx\frac{n_2}{n_1}=\frac{w_2\times M_1}{M_2\times w_1} \)
Here w2 = 20 g, w1 = 100 g of water, M1 = 18 g mol−1 and the relative lowering is 0.1:
\( \displaystyle 0.1=\frac{20\times18}{M_2\times100} \)
\( \displaystyle M_2=\frac{360}{100\times0.1}=\frac{360}{10} \)
\( \displaystyle M_2=36\ \mathrm{g\,mol^{-1}} \)
So the molar mass of the solute is 36 g mol−1.