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Q.

Equivalent conductance at infinite dilution of $B a C l_{2}$ , $H_{2} S O_{4}$ and HCl aqueous solution are $x_{1}, x_{2} a n d x_{3}$ respectively. Equivalent conductance of $B a S O_{4}$ solution is

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a

$x_{1} + x_{2} - 2 x_{3}$

b

$x_{1} - x_{2} - x_{3}$

c

$x_{1} + x_{2} - x_{3}$

d

$x_{1} - 2 x_{2} + x_{3}$

answer is C.

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Detailed Solution

$Λ_{e q}^{0} B a S O_{4} = Λ_{e q}^{0} B a C l_{2} + Λ_{e q}^{0} H_{2} S O_{4} - 2 \times Λ_{e q}^{0} H C l$
$\begin{array}{l} Λ_{M} B a S O_{4} = 2 x_{1} + 2 x_{2} - 2 x_{3} \\ Λ_{M} B a S O_{4} = 2 (x_{1} + x_{2} - x_{3}) \\ Λ_{M} B a S O_{4} = Λ_{e q}^{0} B a S O_{4} \times n - f a c t o r \end{array}$
$Λ_{e q}^{0} B a S O_{4} = \frac{Λ_{M} B a S O_{4}}{n - f a c t o r} = x_{1} + x_{2} - x_{3}$

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