Question

A solute 'S' undergoes a reversible trimerization when dissolved in a certain solvent. The boiling point elevation of its 0.1 molal solution was found to be identical to the boiling point elevation in case of a 0.08 molal solution of a solute which neither undergoes association nor dissociation. To what percent had the solute 'S' undergone trimerization?

• A. $30\%$
• B. $40\%$
• C. $50\%$
• D. $60\%$

Answer 1

Answer: (A)

$30\%$

The trimerization reaction is $3S\rightleftharpoons S_3$
Initially the number of moles of S and $S_3$ are 1 and 0 respectively.
After trimerization, the number of moles of S and $S_3$ are $1-\alpha$ and $\frac{\alpha }{3}$ respectively.
Hence, $i=1-\frac{2\alpha }{3}$
The elevation in boiling point is given by the expression $\Delta T_b=i{K}_{b}m$
For 0.1 molal solution $\Delta T_b=(1-\frac{2\alpha }{3})\times K_b\times 0.1$
For 0.08 molal solution $\Delta T_b=1\times K_b\times 0.08$
Hence, $0.1(1-\frac{2\alpha }{3})=0.08$ or $\alpha =0.3$
Thus, the solute S is $30\%$ trimerized.