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Degree of dissociation of `0.1 N CH_(3)COOH` is (Dissociation constant `= 1 xx 10^(-5)`)A. `10^(-5)`B. `10^(-4)`C. `10^(-3)`D. `10^(-2)` |
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Answer» Correct Answer - D Degree of dissociation `alpha = ?` Normality of solution `= 0.1 N = (1)/(10) N` Dissociation constant `K = 1 xx 10^(-5)` `K = C alpha^(2), alpha = sqrt((K)/(C)) = sqrt((1 xx 10^(-5))/(0.1)) , alpha = 1 xx 10^(-2)`. |
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