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

The Correct Option is A

Explanation:
For solute $S$$3\mathrm{S}\rightleftharpoons {\mathrm{S}}_3$10(Initially)$1-a$$\frac{\alpha }{3}$(at equilibrium)so vant hoff factor for solute s, $i_1=1-\alpha +\frac{\alpha }{3}=1-\frac{2\alpha }{3}$for other solute$i_2=1(\text{as neither assosiacte nor dissociate})$Now elevation of boiling point in both solution is same so molality of first solvent $x$ vant hoff factor for first solvent = molality of second solvent $x$ vant hoff factor for second solvent$\Rightarrow 0.1i_1=0.08i_2$$\Rightarrow 0.1(1-\frac{2\alpha }{3})=0.08$$\Rightarrow \alpha =0.3$$\Rightarrow 30\mathrm{\%}$ trimerization