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CBSE Class 12 Chemistry 2019 Outside Delhi Set 2 Solved Paper

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Question : 18
Total: 27
A solution containing 1.9g per 100‌mL of KCI(M= 74.5gmol−1) is isotonic with a solution containing 3g per 100‌mL of urea (M=60gmol−1). Calculate the degree of dissociation of KCI solution. Assume that both the solutions have same temperature.
Solution:  
π1(‌ urea ‌)‌=π2(KCl)
C1‌RT‌=iC2‌RT
‌
n1
V1
‌=i‌
n2
V2
‌‌(V1=V2)
‌
30
60
‌=i×‌
1.9
74.5

i‌=1.96
α‌=‌
i−1
n−1

‌=‌
1.96−1
2−1

‌=0.96‌ or ‌96%
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