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The measured resistance of a conductance cell containing 7.5 x 10^–3 M solution of KCl at 25° was 1005 ohms. Calculate (a) specific conductance (b) Molar conductance of the solution. Cell constant = 1.25 cm^–1.

Solution

M = 7.5 \times 10^{- 3} R = 1005 ohms .

(a) Specific conductance (K)

= \frac{1}{R} \times \frac{1}{a} where \frac{1}{a} is cell constant .

K = \frac{1}{1005 Ohm} \times 1.25 {cm}^{- 1} K = 1.2437 \times 10^{- 3} {ohm}^{- 1} {cm}^{- 1} = 1.2437 \times 10^{- 3} S {cm}^{- 1}

(b)

Λ_{m} = \frac{1000 \times K}{M} = \frac{1000 \times 1.2437}{7.5 \times 10^{- 3} m} = \frac{1.2437 \times 1000}{7.5} = \frac{1.2437}{7.5} = 165.826 Ω^{- 1} {cm}^{2} {mol}^{- 1} Λ_{m} = 165.826 S {cm}^{2} {mol}^{- 1} .

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