Resistance of 0.2 M solution of an electrolyte is 50 Ω. The specific conductance of the solution is 1.4 S m–1. The resistance of 0.5 M solution of the same electrolyte is 280 Ω. The molar conductivity of 0.5 M solution of the electrolyte in S m2 mol

Engineering

Chemistry

Electrolytic Conductance

Question

Resistance of 0.2 M solution of an electrolyte is 50 Ω. The specific conductance of the solution is 1.4 S m^–1. The resistance of 0.5 M solution of the same electrolyte is 280 Ω. The molar conductivity of 0.5 M solution of the electrolyte in S m² mol^–1 is

5 × 10²

5 × 10³

5 × 10^–4

5 × 10^–3

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K = 4.4*

For 0.2 M-solution $1.4 = \frac{1}{50} \times 4^{*}$ : ...(1)

For 0.5 M-solution: K = $\frac{1}{280}$ × 4* ...(2)

From (1) / (2), k = Sm^–1

Now,

$λ_{m} = \frac{\frac{1}{4} S m^{- 1}}{0 .5 \times 10^{3} moℓ m^{- 3}}$

= 5 × 10^–4 S m² mol^–1

Concentration = 0.2

R_solution = 50 ohm

K = 1.4 S m^–1

Concentration = 0.5

R_solution= 280 ohm

$K = \frac{1}{Resistance}$ cell constant

$\frac{1.4}{100} = \frac{1}{50}$ × cell constant

0.7 = cell constant

$λ_{m} = \frac{(\frac{1}{280} \times 0 .7) \times 1000}{0 .5}$

= $\frac{700 \times 2}{280}$ = S cm²/mΩ

= 5 × 10^–4 S m²/mΩ