At 250C the molar conductances at infinite dilution for the strong electrolytes NaOH, NaCl and BaCl2 are 248×10−4,126×10−4 and 280×10−4sm2mol−1 respectively.^m0BaOH2 in sm2mol-1 is

^m0BaOH2=^m0BaCl2+2^m0NaOH−2^m0NaCl =(280+2×248-2×126)×10-4 =524 ×10-4

At 250C the molar conductances at infinite dilution for the strong electrolytes NaOH, NaCl and BaCl2 are 248×10−4,126×10−4 and 280×10−4sm2mol−1 respectively.^m0BaOH2 in sm2mol-1 is

^m0BaOH2=^m0BaCl2+2^m0NaOH−2^m0NaCl =(280+2×248-2×126)×10-4 =524 ×10-4

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

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At 250C the molar conductances at infinite dilution for the strong electrolytes NaOH, NaCl and BaCl2 are 248×10−4,126×10−4 and 280×10−4sm2mol−1 respectively.^m0BaOH2 in sm2mol-1 is

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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