The molar conductance's at infinite dilution of BaCl2, NaCl and NaOH are respectively 280×10−4,126.5×10−4,248×10−4Sm2mol−1The molar conductance at infinite dilution for Ba(OH)2 is

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The molar conductance's at infinite dilution of BaCl₂, NaCl and NaOH are respectively
$280 \times 10^{- 4}, 126 .5 \times 10^{- 4}, 248 \times 10^{- 4} {S m}^{2} {m o l}^{- 1}$ The molar conductance at infinite dilution for Ba(OH)₂ is

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$523 \times 10^{- 4} {Sm}^{2} {mol}^{- 1}$

$65 \times 10^{- 4} {Sm}^{2} {mol}^{- 1}$

$52 .3 \times 10^{- 4} {Sm}^{2} {mol}^{- 1}$

$5.23 \times 10^{- 4} {Sm}^{2} {mol}^{- 1}$

answer is A.

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Detailed Solution

$μ_{{BaCl}_{2}}^{\infty} = μ_{{Ba}^{2 +}}^{\infty} + 2 μ_{C 1^{-}}^{\infty}$ ...(i) $μ_{NaC 1}^{\infty} = μ_{{Na}^{+}}^{\infty} + μ_{{Cl}^{-}}^{\infty}$ ....(ii) $μ_{NaOH}^{\infty} = μ_{{Na}^{+}}^{\infty} + μ_{{HO}^{-}}^{\infty}$ .....(iii)for $Ba (OH)_{2} (i) + 2 (iii) - 2 (ii) \Rightarrow μ_{(Ba (OH)_{2})}$ $= (280 \times 10^{- 4}) + 2 (248 \times 10^{- 4}) - 2 (126 .5 \times 10^{- 4})$ $= 523 \times 10^{- 4} {Sm}^{2} {mol}^{- 1}$