If the cell constant is $0.40 c m^{- 1}$ , the conductivity of 0.051 M NaCl solution having R = 1850 ohm is equal to:

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

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

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

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5.04 \times 10^{- 5} S c m^{- 1}

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Solution

The correct option is B $2.16 \times 10^{- 4} S, c m^{- 1}$
$κ = G \times G^{}$
where $κ = c o n d u c t i v i t y G = c o n d u c t a n c e = \frac{1}{1850} = 0.00054 S G^{} = c e l l c o n s t a n t = 0.4 c m^{- 1} κ = 0.4 \times 0.00054 = 2.16 \times 10^{- 4} S c m^{- 1}$

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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 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})

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 $C a S O_{4}$ .