Molar conductance at infinite dilution of BaCl2 , NaOH and NaCl are ∧1 , ∧2 and ∧3 respectively. Molar conductance of BaOH2 would be (all measurements in CGS units)

2 ( ∧ 1 + 2 ∧ 2 − 2 ∧ 3 )

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Molar conductance at infinite dilution of $B a C l_{2}$ , $N a O H$ and NaCl are $\land_{1}$ , $\land_{2}$ and $\land_{3}$ respectively. Molar conductance of $B a {(O H)}_{2}$ would be (all measurements in CGS units)

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$2 (\land_{1} + 2 \land_{2} - 2 \land_{3})$

$\frac{\land_{1} + 2 \land_{2} - 2 \land_{3}}{2}$

$\land_{1} + \land_{2} - \land_{3}$

$\land_{1} + 2 \land_{2} - 2 \land_{3}$

answer is B.

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$\land_{1} = \land_{B a^{2 +}} + 2 \land_{{Cl}^{-}}$ .....( i)

$\land_{2} = \land_{{Na}^{+}} + \land_{{OH}^{-}}$ (ii)

$\land_{3} = \land_{{Na}^{+}} + \land_{{Cl}^{-}} (i i i)$

$(i) + (i i) \times 2 - (i i i) \times 2$

$\land_{1} + 2 \land_{2} - 2 \land_{3} = \land_{{Ba}^{2 +}} + 2 \land_{{OH}^{-}}$

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