At 25⁰C the molar conductances at infinite dilution for the strong electrolytes NaOH, NaCl and BaCl₂ are $248 \times 10^{- 4}, 126 \times 10^{- 4}$ and $280 \times 10^{- 4} s m^{2} m o l^{- 1}$ respectively. $^_{m}^{0} B a {(O H)}_{2}$ in sm²mol^-1 is

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$52.4 \times 10^{- 4}$

$524 \times 10^{- 4}$

$402 \times 10^{- 4}$

$262 \times 10^{- 4}$

answer is B.

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Detailed Solution

$^_{m}^{0} B a {(O H)}_{2} =^_{m}^{0} B a C l_{2} + 2^_{m}^{0} N a O H - 2^_{m}^{0} N a C l = (280 + 2 \times 248 - 2 \times 126) \times 10^{- 4} = 524 \times 10^{- 4}$

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