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 $1.29 S m^{-1}.$. 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. $124\times10^{-4} S m^{2} mol^{-1}$
• B. $1240\times10^{-4} S m^{2} mol^{-1}$
• C. $1.24\times10^{-4} S m^{2} mol^{-1}$
• D. $12.4\times10^{-4} S m^{2} mol^{-1}$

Answer 1

Answer: (D)

$R=100 \Omega$
$K=\frac{1}{R} \left(\frac{1}{a}\right)$
$\frac{1}{a} \left(cell \,constant\right)=1.29\times100 m^{-1}$
$Given \, R=520 \Omega ; C=0.2 M$
$\mu$ (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 $\mu=\frac{1}{520} \times129 \times\frac{1000}{0.2}\times10^{-6} m^{3}$
$=12.4\times10^{-4} Sm^{2} mol^{-1}$