Following limiting molar conductivities are given below
$λ_{m}^{0} (H_{2} S O_{4}) = x S c m^{2} m o l^{- 1}$
$λ_{m}^{0} (K_{2} S O_{4}) = y S c m^{2} m o l^{- 1}$
$λ_{m}^{0} (C H_{3} C O O K) = z S c m^{2} m o l^{- 1}$
$λ_{m}^{0} (i n S c m^{2} m o l^{- 1})$ for $C H_{3} C O O H$ will be ?

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$\frac{(x - y)}{2} + z$

$x - y + z$

$x - y + 2 z$

$x + y - z$

answer is D.

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Detailed Solution

$2 C H_{3} C O O K + H_{2} S O_{4} \to 2 C H_{3} C O O H + K_{2} S O_{4}$
In question given, that it is asked only for one mole of $C H_{3} C O O H$
$∴$ the above equation is written as
Question Image

$∴ λ_{m}^{0} C H_{3} C O O H = z + \frac{x}{2} - \frac{y}{2}$

or $z + \frac{(x - y)}{2}$

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