Question

Resistance of a conductivity cell filled with a solution of an electrolyte of concentration 0.1 M is 100 $\Omega$. The conductivity of this solution is $1.29S{m}^{-1}.$. Resistance of the same cell when filled with 0.2 M of the same solution is 520 $\Omega$. The molar conductivity of 0.02 M solution of the electrolyte will be:

• A. $124\times 10^{-4}S{m}^{2}mo{l}^{-1}$
• B. $1240\times 10^{-4}S{m}^{2}mo{l}^{-1}$
• C. $1.24\times 10^{-4}S{m}^{2}mo{l}^{-1}$
• D. $12.4\times 10^{-4}S{m}^{2}mo{l}^{-1}$

Answer 1

Answer: (D)

$R=100\Omega$
$K=\frac{1}{R}\left(\frac{1}{a}\right)$
$\frac{1}{a}\left(cellconstant\right)=1.29\times 100m^{-1}$
$GivenR=520\Omega ;C=0.2M$
$\mu$ (molar conductivity) = ?
$\mu =K\times V$
(K can be calculated as $k=\frac{1}{R}\left(\frac{1}{a}\right)$ now cell constant is known)
Hence $\mu =\frac{1}{520}\times 129\times \frac{1000}{0.2}\times 10^{-6}{m}^3$
$=12.4\times 10^{-4}S{m}^{2}mo{l}^{-1}$