Following limiting molar conductivities are given as: [2019]

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Solution

Q.
Following limiting molar conductivities are given as:

$λ_{m (H_{2} {SO}_{4})}^{\circ} = x$ $S$ ${cm}^{2}$ ${mol}^{- 1}$

$λ_{m (K_{2} {SO}_{4})}^{\circ} = y$ $S$ ${cm}^{2}$ ${mol}^{- 1}$

$λ_{m ({CH}_{3} COOK)}^{\circ} = z$ $S$ ${cm}^{2}$ ${mol}^{- 1}$

$λ_{m}^{\circ} (in$ $S$ ${cm}^{2}$ ${mol}^{- 1})$ ${for CH}_{3} COOH will be$ [2019]

1 $x - y + 2 z$

2 $x + y - z$

3 $x - y + z$

4 $\frac{(x - y)}{2} + z$

Ans.
(4)

According to Kohlrausch's law,

$Λ_{m}^{o} {for CH}_{3} COOH = λ_{{CH}_{3} {COO}^{-}}^{\circ} + λ_{H^{+}}^{\circ}$

$Λ ° {for H}_{2} {SO}_{4} = 2 λ_{H^{+}}^{\circ} + λ_{{SO}_{4}^{2 -}}^{\circ} = x$ $S$ ${cm}^{2}$ ${mol}^{- 1}$ ...(i)

$Λ ° {for K}_{2} {SO}_{4} = 2 λ_{K^{+}}^{\circ} + λ_{{SO}_{4}^{2 -}}^{\circ} = y$ $S$ ${cm}^{2}$ ${mol}^{- 1}$ ...(ii)

$Λ ° {for CH}_{3} COOK = λ_{{CH}_{3} {COO}^{-}}^{\circ} + λ_{K^{+}}^{\circ} = z$ $S$ ${cm}^{2}$ ${mol}^{- 1}$ ...(iii)

On adding equation (i) and 2 $\times$ (iii) and subtracting (ii), we get

$2 λ_{H^{+}}^{\circ} + λ_{{SO}_{4}^{2 -}}^{\circ} + 2 λ_{{CH}_{3} {COO}^{-}}^{\circ} + 2 λ_{K^{+}}^{\circ} - 2 λ_{K^{+}}^{\circ} - λ_{{SO}_{4}^{2 -}}^{\circ} = x + 2 z - y$

$2 λ_{H^{+}}^{\circ} + 2 λ_{{CH}_{3} {COO}^{-}}^{\circ} = x + 2 z - y$

$λ_{H^{+}}^{\circ} + λ_{{CH}_{3} {COO}^{-}}^{\circ} = \frac{(x - y)}{2} + z$

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