A dilute solution contains ‘t’ moles of solute X in 1 kg of solvent with molal elevation constant $K_{b}$ , the solute dimerises in the solution according to the following equation. The degree of association is $α . 2 X ⇌ X_{2}$

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

The colligative properties observed will be

${(T_{b})}_{o b s} < {(T_{b})}_{t h e o r e t i c a l}; {(T_{f})}_{o b s} < {(T_{f})}_{t h e o r e t i c a l}; Δ P_{o b s} < Δ P_{t h e o r e t i c a l}$

The equilibrium constant for the process can be expressed as: $K = \frac{K_{b} (K_{b} t - Δ T_{b})}{{[2 Δ T_{b} - K_{b} t]}^{2}}$

The molecular mass observed will be lesser than actual molecular mass

The degree of association $α = \frac{2 (K_{b} t - Δ T_{b})}{K_{b} T}$

answer is A, D.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

$\underset{1 - α}{2 X} ⇌ \underset{\frac{α}{2}}{X_{2}} i = 1 - \frac{α}{2} \frac{{(Δ t_{b})}_{0.65}}{K_{b} t} = i = 1 - \frac{α}{2} \frac{α}{2} = 1 - \frac{Δ t_{b}}{K_{b} t} = \frac{K_{b} t - Δ t_{b}}{K_{b} t} (A) α = \frac{2 (K_{b} t - Δ t_{b})}{K_{b} t}$