Following limiting molar conductivities are given as λm0H2SO4=x S cm2 mol−1λm0K2SO4=y S cm2 mol−1λm0CH3COOK=z S cm2 mol−1λm0in S cm2 mol−1 for CH3COOH will be

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Following limiting molar conductivities are given as
$λ_{m}^{0} (H_{2} S O_{4}) = x S c m^{2} m o l^{- 1}$
$λ_{m}^{0} (K_{2} S O_{4}) = y S c m^{2} m o l^{- 1}$
$λ_{m}^{0} (C H_{3} C O O K) = z S c m^{2} m o l^{- 1}$
$λ_{m}^{0} (i n S c m^{2} m o l^{- 1}) f o r C H_{3} C O O H w i l l b e$

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

$x + y + z$

$x - y + z$

$x - y + 2 z$

answer is A.

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Detailed Solution

$C H_{3} C O O H + K O H \to C H_{3} C O O K + H_{2} O$
$\begin{array}{l} 2 K O H + H_{2} S O_{4} \to K_{2} S O_{4} + 2 H_{2} O \\ C H_{3} C O O H = \frac{1}{2} H_{2} S O_{4} - \frac{1}{2} K_{2} S O_{4} + C H_{3} C O O H \\ = \frac{x - y}{2} + z \end{array}$