The resistance of a conductance cell, when filled with a 0.02 mol dm^-3 solution of KCI (k = sp. Conductivity = 2.768 × 10^-3 ohm^-1 cm^-1) was found to be 457 ohms. When the same cell was filled with a solution of CaCl₂, the measured resistance was 1050 ohms. Incorrect statements among the following are

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$equivalent conductance of 0.02 M {CaCl}_{2} is (in {Sm}^{2} {eq}^{- 1}), 1 \times 10^{- 3}$

$specific conductance of 0 .02 M {CaCl}_{2} is (in {Sm}^{- 1}) 0 .240$

$molar conductance of 0 .02 M {CaCl}_{2} is (in {Sm}^{2} {mol}^{- 1}) is, 6 \times 10^{- 3}$

cell constant of the conductance cell is (in m^-1) 1.265

answer is A, B, C.

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Detailed Solution

(1) $k = \frac{l}{a R}; \frac{l}{a} = k \times R = 2.768 \times 10^{- 3} {ohm}^{- 1} \times 100 m^{- 1} \times 457 = 126.5 m^{- 1}$

(2) $k (specific conductivity) of {CaCl}_{2} = \frac{l}{a} \times \frac{1}{R} = 126.5 m^{- 1} \times \frac{1}{1050}$

$= 0.120 {Sm}^{- 1}$

(3) $\land (Equivalent conductivity) of {CaCl}_{2} = \frac{k}{c} = \frac{0.120 {Sm}^{- 1}}{40 {eqm}^{- 3}} = 3 \times 10^{- 3} {Sm}^{2} {eq}^{- 1}$

(4) $μ (molarconductance) of {CaCl}_{2} = \frac{k}{c^{'}} = \frac{0.120}{20}$

$= 6 \times 10^{- 3} {Sm}^{2} {mol}^{- 1} (c^{'} = {moldm}^{- 3})$

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