The resistance and conductivity of 0.02 M KCl solution are 82.4 ohm and $0.002768 S c m^{- 1}$ , respectively. When filled with $0.005 N K_{2} S O_{4}$ , the solution had a resistance of 324 ohm. Calculate conductivity of $K_{2} S O_{4}$ solution (conductance and cell constant calculated in previous problems are $3.086 \times 10^{(- 3)} S$ and $0.228 c m^{- 1})$

3.04 \times 10^{- 4} S c m^{- 1}

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7.04 \times 10^{- 4} S c m^{- 1}

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5.04 \times 10^{- 4} S c m^{- 1}

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9.04 \times 10^{- 4} S c m^{- 1}

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Solution

The correct option is B $7.04 \times 10^{- 4} S c m^{- 1}$
Conductivity or specific conductance $k = G \times \frac{1}{a} = 3.086 \times 10^{- 3} \times 0.2281 = 7.04 \times 10^{- 4} S c m^{- 1}$

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Q. The resistance and conductivity of 0.02 M KCl solution are 82.4 ohm and

0.002768 S c m^{- 1}

, respectively. When filled with

0.005 N K_{2} S O_{4}

, the solution had a resistance of 324 ohm. Calculate equivalent conductivity of

K_{2} S O_{4}

solution.

The conductivity of $0.001 M$ $N a_{2} S O_{4}$ solution is $2.6 \times 10^{- 4} {S cm}^{- 1}$ and increases to $7 \times 10^{- 4} {S cm}^{- 1}$ , when the solution is saturated with $C a S O_{4}$ . The molar conductivity of $N a^{+}$ and $C a^{2 +}$ are $50$ and $120 {S cm}^{2} {mol}^{- 1}$ , respectively. Neglect the conductivity of water, find the solubility product of $C a S O_{4}$ .