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 Sm–1. Resistance of the same cell when filled with 0.02 M of the same solution is 520 $\Omega$. The molar conductivity of 0.02 M solution of the electrolyte will be : (Take$\frac{129}{520}$ = 0.248)

• A. $\text{124 × 1}\text{0}^{\text{-4}}\text{ S}\text{m}^{2}\text{mo}\text{l}^{\text{-1}}$
• B. $\text{1240 }\text{×}\text{ 1}\text{0}^{\text{-4 }}\text{S}\text{m}^{2}\text{mo}\text{l}^{\text{-1}}$
• C. $\text{1.24 S}\text{m}^{2}\text{mo}\text{l}^{\text{-1}}$
• D. $\text{12.4 }\text{×}\text{ 1}\text{0}^{\text{-4}}\text{ S}\text{m}^{2}\text{mo}\text{l}^{\text{-1}}$

Answer 1

Answer: (D)

$\text{12.4 }\text{×}\text{ 1}\text{0}^{\text{-4}}\text{ S}\text{m}^{2}\text{mo}\text{l}^{\text{-1}}$

Answer 2

$\text{C = 0.1 M,} \text{R = 100}\Omega$

$\text{K = 1.29 S}\text{m}^{\text{-1}}\text{ = }\frac{1}{\text{100}}\text{ }\text{×}\frac{\mathcal{l}}{A}\text{ .}$

$\text{0.02 M,} \text{R = 520}\Omega.\text{K = }\frac{1}{\text{520}}\text{×}\text{ 129}$

$Å_{M}\text{ =}\frac{\frac{1}{\text{520}} \times 129}{1000 \times 0.02}\text{= 124 }\text{×}\text{ 1}\text{0}^{\text{-4}}\text{ S}\text{m}^{2}\text{mo}\text{l}^{\text{-1}}.$ s = 1.29 × 100

= 129.

For O the Contraption r = 129.

$K' = \frac{\sigma}{R} = \frac{129}{520}; \Lambda_{M} = \frac{1000K}{M} = \frac{1000 \times 129}{0.02 \times 500} = 5Cm^{2}mol^{- 1}.$