Following limiting molar conductivities are given as _ m (H_ 2 SO_ 4 ) ^ 0 =x ;S ; cm^ 2 mol^ -1 _ m (K_ 2 SO_ 4 ) ^ 0 =y ;S ;cm^ 2 mol^ -1 _ m(CH_3 COOK) ^ 0 =z ; S ;cm^ 2 mol^ -1 _ m ^ 0 (in ; S ;cm^ 2 mol^ -1 ) for CH_ 3 COOH will be

Following limiting molar conductivities are given as _ m (H_ 2 SO…

MCQ

Following limiting molar conductivities are given as

$\lambda_{\mathrm{m}\left(\mathrm{H}_{2} \mathrm{SO}_{4}\right)}^{0}=\mathrm{x} \;\mathrm{S}\; \mathrm{cm}^{2} \mathrm{mol}^{-1}$

$\lambda_{\mathrm{m}\left(\mathrm{K}_{2} \mathrm{SO}_{4}\right)}^{0}=\mathrm{y} \;\mathrm{S\;cm}^{2} \mathrm{mol}^{-1}$

$\lambda_{\mathrm{m}(\mathrm{CH_3} \mathrm{COOK})}^{0}=\mathrm{z}\; \mathrm{S\;cm}^{2} \mathrm{mol}^{-1}$

$\lambda_{\mathrm{m}}^{0}\left(\mathrm{in}\; \mathrm{S} \;\mathrm{cm}^{2} \mathrm{mol}^{-1}\right)$ for $\mathrm{CH}_{3} \mathrm{COOH}$ will be

A
$x + y + 2 z$
B
$x + y - z$
C
$x + y + z$
✓
$\frac{x-y}{2}+z$

✓

Answer

Correct option: D.

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

d
$\mathrm{CH}_{3} \mathrm{COOH} \rightarrow \mathrm{CH}_{3} \mathrm{COO}^{-}+\mathrm{H}^{+}\dots (1)$

$\mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow 2 \mathrm{H}^{+}+\mathrm{SO}_{4}^{-2}\dots (2)$

$\mathrm{K}_{2} \mathrm{SO}_{4} \rightarrow 2 \mathrm{K}^{+}+\mathrm{SO}_{4}^{-2}\dots (3)$

$\mathrm{CH}_{3} \mathrm{COOK} \rightarrow \mathrm{CH}_{3} \mathrm{COO}^{-}+\mathrm{K}^{+}\dots (4)$

According to Kohlrausch's law

$\lambda_{\mathrm{CH}_{3} \mathrm{COOH}}^{\circ}=\lambda_{\mathrm{CH}_{3} \mathrm{COO}^{-}}^{\circ}+\lambda_{\mathrm{H}^{+}}^{\circ}$

eq. $(1)=$ eq. $(4)+$ eq. $\frac{(2)}{2}-$ eq. $\frac{(3)}{2}$

$\therefore \quad \lambda_{\mathrm{CH}_{3} \mathrm{COOH}}^{\circ}=z+\frac{\mathrm{x}}{2}-\frac{\mathrm{y}}{2}$

$\lambda_{\mathrm{CH}_{3} \mathrm{COOH}}^{\circ}=\frac{(\mathrm{x}-\mathrm{y})}{2}+\mathrm{z}\left(\mathrm{S} \times \mathrm{cm}^{2} \mathrm{mol}^{-1}\right)$

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