A non-volatile solute HX undergo complete (100%) ionization in water at any temperature above boiling point of pure water. The same solute HX undergo complete (100%) dimerization in water at any temperature below freezing point of pure water. The sum of values of both colligative properties elevation in boiling point and depression in freezing point for 1.0 molal aqueous solution of HX is (Round off answer to nearest integer)
$[K_{b} = 0.52, K . k g . m o l e^{- 1}, K_{f} = 1.86 K . k g . m o l e^{- 1}]$

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

answer is 2.

(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

$Δ T_{f} = (1.86) (\frac{1}{2}) = 0.93 K$

$Δ T_{b} = (0.52) (2) = 1.04 K$

$Δ T_{b} + Δ T_{f} = 1.97 ≃ 2$

Watch 3-min video & get full concept clarity

JEE

NEET

Foundation JEE

Foundation NEET

CBSE