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CBSE Class 12 Chemistry 2019 Outside Delhi Set 2 Solved Paper

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Question : 18 of 27
Marks: +1, -0
A solution containing 1.9 g1.9\text{ g} per 100 mL100\text{ mL} of KCI(M=\mathrm{KCI}(M= 74.5 g mol−1)74.5\text{ g}\text{ mol}^{-1}) is isotonic with a solution containing 3 g3\text{ g} per 100 mL100\text{ mL} of urea (M=60 g mol−1)(M=60\text{ g}\text{ mol}^{-1}). Calculate the degree of dissociation of KCI\mathrm{KCI} solution. Assume that both the solutions have same temperature.
Solution:  
π1( urea )=π2(KCl)\pi_1(\text{ urea })=\pi_2(\mathrm{KCl})
C1RT=iC2RTC_1 RT=i C_2 RT
n1V1=in2V2(V1=V2)\frac{n_1}{V_1}=i\frac{n_2}{V_2}\quad(V_1=V_2)
3060=i×1.974.5\frac{30}{60}=i\times\frac{1.9}{74.5}
i=1.96i=1.96
α=i−1n−1\alpha=\frac{i-1}{n-1}
=1.96−12−1=\frac{1.96-1}{2-1}
=0.96 or 96%=0.96\text{ or }96\%
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