Question

Dry $N_2$ was passed successively through a solution of solute $A_{2}B$ in water at ${27}^{o}C$ and then through pure water. The loss in mass of solution was 2.5 g and that of pure water was 0.04 g. Find the degree of dissociation of salt $A_{2}B$ when solution of $A_{2}B$ has same concentration as 1.86% (mass/vol) solution of urea at ${27}^{o}C$. Assume molarity is equal to molality. Divide answer by 10 and write the nearest integer.

Answer 1

Molar mass of urea is 60 g/mol.

The concentration of solution is 1.86% (mass/vol.)

1000 g of solution contain 18.6 g or $\frac{18.6}{60}=0.31$ moles of urea or 0.31 moles (n) of salt $A_{2}B$

Moles of water (N) in the solution $=\frac{1000-18.6}{18}=54.5$ moles

$\frac{P^0-P^s}{P^0}=\frac{\text{Loss in weight of water}}{\text{Loss in weight of solution}}=\frac{i\times n}{N}$

Here i is the van't Hoff factor.

$\frac{0.04}{2.5}=\frac{i\times 0.31}{54.5}$

$i=2.8$

Total number of ions produced by dissociation of 1 molecule of A2B is $n^{\prime }=3$

The degree of dissociation $\alpha =\frac{i-1}{n^{\prime }-1}=\frac{2.8-1}{3-1}=0.907$ or $90.7\%$

Hence, answer is 9.