Question

Each of the following compounds has been dissolved in water to make its 0.001 M solution. Rank them in order of their increasing conductivity in solution (assume 100% ionization in each case).

(a) $KCl$ (b) $K_{2}S{O}_4$ (c) $C{H}_{3}COOH$

• A. $c<a<b$
• B. $c<b<a$
• C. $a<b<c$
• D. $a=b=c$

Answer 1

Answer: (A)

$c<a<b$

Answer:- (A) $c<a<b$

$KCl\rightleftharpoons K^{+}+{Cl}^{-}$

$K_{2}S{O}_4\rightleftharpoons 2K^{+}+{S{O}_4}^{-2}$

$C{H}_{3}COOH\rightleftharpoons {C{H}_{3}COO}^{-}+H^{+}$

The conducticity of the solution increases as the number of ions increases. Number of free ions in $KCl,K_{2}S{O}_4$ and $C{H}_{3}COOH$ are $2,3$ and $2$ respectively.

Since $C{H}_{3}COOH$ is weak electrolyte than $KCl$, thus the conductivity of $KCl$ will be more than $C{H}_{3}COOH$.

Hence the increasing order of conductivity is-

$\left(c\right)<\left(a\right)<\left(b\right)$