CBSE Class 12 Chemistry 2019 Outside Delhi Set 2 Solved Paper

Question : 18 of 27

Marks: +1, -0

A solution containing

1.9 {\g}

per

100 {\mL}

{\KCI}( {\M}=

74.5 {\g} {\mol}^{-1})

is isotonic with a solution containing

3 {\g}

per

100 {\mL}

of urea

( {\M}=60 {\g} {\mol}^{-1})

. Calculate the degree of dissociation of

{\KCI}

solution. Assume that both the solutions have same temperature.

Solution:

π_1(\;\text " urea "\;)\;=π_2( {\KCl})

{\C}_1 {\RT}\;= {\i} {\C}_2 {\RT}

\;{n_1}/{V_1}\;=i \;{n_2}/{V_2} \;\; (V_1=V_2)

\;{30}/{60}\;=i × \;{1.9}/{74.5}

i\;=1.96

α\;=\;{i-1}/{n-1}

\;=\;{1.96-1}/{2-1}

\;=0.96 \;\text " or "\; 96 \%

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