The specific conductivity of a 0.001 M $N a_{2} S O_{4}$ solution is $2.6 \times 10^{- 4} S c m^{- 1}$ and becomes $7.0 \times 10^{- 4} S c m^{- 1}$ when the solution is saturated with $C a S O_{4}$ . The limiting molar conductivities of $N a^{+}$ and $C a^{2 +}$ ions are 50 and $120 S c m^{2} m o l^{- 1}$ respectively. Neglecting specific conductivity of water the correct among the following

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Specific conductivity of $C a S O_{4}$ solution is $4.4 \times 10^{- 7} S c m^{- 1}$

$\land_{m (S O_{4}^{- 2})}^{0} = 160 S c m^{2} m o l^{- 1}$

$K_{s p} o f C a S O_{4} = 4 \times 10^{- 6} M^{2} (n e a r l y)$

$\land_{m (N a_{2} S O_{4})}^{0} = 260 S c m^{2} m o l^{- 1}$

answer is A, B, C.

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Detailed Solution

$a) \Rightarrow \land_{m} (N a_{2} S O_{4}) = \frac{2.6 \times 10^{- 4} \times 1000}{0.001} = 260 \Rightarrow λ (S O_{4}^{2 -}) = \land (N a_{2} S O_{4}) - 2 λ (N a^{+}) = 260 - 2 \times 50 = 160 S c m^{2} m o l^{- 1} \Rightarrow \land (C a S O_{4}) = λ (C a^{2 +}) + λ (S O_{4}^{2 -}) = 120 + 160 = 280$

$\Rightarrow$ Concentration of $C a^{2 +}$ ion is solution $= \frac{k \times 1000}{\land} = \frac{4.4 \times 10^{- 4} \times 1000}{280} = 1.57 \times 10^{- 3} M$