Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

A $0.01\; m$ aqueous solution of $AlCl_{3}$ freezes at $-0.068^{\circ}C$ . Calculate the percentage of dissociation. [Given : $K_{f}$ for Water $=1.86\; K\; kg\; mol^{-1}$ ]

Answers (1)

ΔTf = i(K_f)m

0.068 = i x 1.86 x 0.01

i = 3.65 or 3.656

α = i-1/n-1

α = 0.883 or 0.885

Hence percentage of dissociation is 88.3% or 88.5%

Posted by

Safeer PP

View full answer

Crack CUET with india's "Best Teachers"

HD Video Lectures
Unlimited Mock Tests
Faculty Support

cuet_ads

Similar Questions

Trending Articles/News

anam-khan

CBSE practical exams 2026 will be cancelled over non-compliance with guidelines, says board

Latest Question