For a strong electrolyte,Lambda_{m} increases | vedclass

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(B) The given equation is $\Lambda_{m} = \Lambda_{m}^{\circ} - A\sqrt{c}$. Using the data for $c = 0.04 \ mol \ L^{-1}$ and $\Lambda_{m} = 96.1 \ S \

Vedclass · Accepted Answer

$4$

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(B) The given equation is $\Lambda_{m} = \Lambda_{m}^{\circ} - A\sqrt{c}$. Using the data for $c = 0.04 \ mol \ L^{-1}$ and $\Lambda_{m} = 96.1 \ S \ cm^2 \ mol^{-1}$: $96.1 = \Lambda_{m}^{\circ} - A 	imes \sqrt{0.04} = \Lambda_{m}^{\circ} - 0.2A \quad \dots(I)$ Using the data for $c = 0.09 \ mol \ L^{-1}$ and $\Lambda_{m} = 95.7 \ S \ cm^2 \ mol^{-1}$: $95.7 = \Lambda_{m}^{\circ} - A 	imes \sqrt{0.09} = \Lambda_{m}^{\circ} - 0.3A \quad \dots(II)$ Subtracting equation $(II)$ from equation $(I)$: $(96.1 - 95.7) = (-0.2A) - (-0.3A)$ $0.4 = 0.1A$ $A = \frac{0.4}{0.1} = 4$ Thus,the value of constant $A$ is $4$.

$c \ [mol \ L^{-1}]$	$0.04$	$0.09$	$0.16$	$0.25$
$\Lambda_{m} \ [S \ cm^2 \ mol^{-1}]$	$96.1$	$95.7$	$95.3$	$94.9$

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