SECTION -C A 0.01 {\m} aqueous solution of {\AlCl}

CBSE Class 12 Chemistry 2020 Delhi Set 1 Solved Paper

Question : 28

Total: 37

SECTION -C

A 0.01m aqueous solution of AlCl3 freezes at 0.068∘C. Calculate the percentage of dissociation.
[Given : Kf for Water =1.68K‌kg‌mol−1 ]

Solution:

Given, m=0.01m
‌∆Tf( s)=−0.068∘C
‌Kf(aq)=1.86K‌kg‌mol−1
‌∆Tf=iKfm
‌i=‌

∆Tf

×m
‌i=‌

0.068

1.86

×0.01m=3.65

Total number of moles at equilibrium
=1−α+α+3α=1+3α
l=‌‌

‌ Total no. of moles at equilibrium ‌

‌ Initial no. of moles ‌

‌=‌

1+3α

3.65‌=1+3α
α‌=‌

3.65−1

Percentage dissociation =0.88%.

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