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The equivalent conductivity of a solution containing 2.54g of ${CuSO}_{4}$ per liter is $91.0 Ω^{- 1} {cm}^{2} {eq}^{- 1} .$ Its conductivity would be

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$1.45 \times 10^{- 3} Ω^{- 1} {cm}^{- 1}$

$2.9 \times 10^{- 3} Ω^{- 1} {cm}^{- 1}$

$2.17 \times 10^{- 3} Ω^{- 1} {cm}^{- 1}$

$2.90 \times 10^{- 3} Ω^{- 1} {cm}^{2}$

answer is D.

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Detailed Solution

$N = \frac{2.54}{\frac{159.5}{2}} \times \frac{1}{1} = 0.0159 \times 2 = 0.0318$

$λ = \frac{K \times 1000}{N} \Rightarrow K = \frac{91 \times 0.0159 \times 2}{1000}$

$= 1.45 \times 10^{- 3} \times 2$

$2.9 \times 10^{- 3} Ω^{- 1} {cm}^{- 1}$

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