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CBSE Class 12 Chemistry 2020 Delhi Set 1 Solved Paper

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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
Kf
×m
‌i=‌
0.068
1.86
×0.01m=3.65
‌AlCl3→Al3++3Cl−
‌initial 1‌mol‌0‌0
‌ At equilibrium 1−α‌α‌3α

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
1

3.65‌=1+3α
α‌=‌
3.65−1
3

Percentage dissociation =0.88%.
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