Electrochemistry Result Question 3
3. Consider the statements $\mathrm{S} _1$ and $\mathrm{S} _2$ :
$S _1$ : Conductivity always increases with decrease in the concentration of electrolyte.
$\mathbf{S} _2$ : Molar conductivity always increases with decrease in the concentration of electrolyte.
The correct option among the following is
(a) $\mathrm{S} _1$ is correct and $\mathrm{S} _2$ is wrong
(b) $\mathrm{S} _1$ is wrong and $\mathrm{S} _2$ is correct
(c) Both $S _1$ and $S _2$ are wrong
(d) Both $\mathrm{S} _1$ and $\mathrm{S} _2$ are correct
(2019 Main, 10 April I)
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Answer:
Correct Answer: 3. ( b )
Solution: 3. The explanation of statements $\left(\mathrm{S}_1\right.$ and $\left.\mathrm{S}_2\right)$ are as follows :
In conductivity cell, conductivity $(\kappa)$ is equal to the sum of ionic conductances $(c)$, of an electrolytic solution present is unit volume of the solution enclosed by two electrodes of unit area $(a \neq 1)$ separated by a unit length $(l=1)$.
$ \kappa=c \times \frac{l}{a} \Rightarrow \kappa=c \text { when } l=1, a=1 $
So, with decrease in the concentration of electrolyte, number of ions in the given unit volume also decreases, i.e. $\kappa$ [conductivity] also decreases.
Thus, statement $\mathrm{S}_1$ is wrong. $\mathrm{S}_2$ : Molar conductivity $\left(\lambda_m\right)$ is defined as the conducting power of all the ions present in a solution containing 1 mole of an electrolyte.
$ \lambda_m=\kappa \times V_{\mathrm{mL}}=\kappa \times \frac{1000}{M} $
where, $V_{\mathrm{mL}}=$ volume in mL containing 1 mole of electrolyte $m=$ molar concentration ( $\mathrm{mol} / \mathrm{L}$ )
So, in a conductivity cell
$ \lambda_m \propto \frac{1}{M} $
i.e. molar conductivity increases with decrease in the concentration $(M)$ of electrolyte. Thus, statement $\mathrm{S}_2$ is correct.
BETA
