Electrochemistry - Result Question 1
####1. Following limiting molar conductivities are given as
$\lambda_m^{\circ}(H_2 SO_4)=x S cm^{2} mol^{-1}$
[NEET Odisha 2019]
$\lambda_m^{\circ}(K_2 SO_4)=y S cm^{2} mol^{-1}$
$\lambda_m^{\circ}(CH_3 COOK)=z S cm^{2} mol^{-1}$
$\lambda_m^{\circ}$ (in $S cm^{2} mol^{-1}$ ) for $CH_3 COOH$ will be
(a) $\frac{(x-y)}{2}+z$
(b) $x-y+2 z$
(c) $x+y+z$
(d) $x-y+z$
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Solution:
- (a) According to Kohlrausch’s law
$ \lambda_m^{\circ}(AB)=\lambda_m^{\circ}(A^{+})+\lambda_m^{\circ}(B-) $
So, $\lambda_m^{\circ}(CH_3 COOH)=\lambda_m^{\circ}(CH_3 COO^{-})+\lambda_m^{\circ}(H^{+})$ So $\lambda_m^{\circ}(CH_3 COOH)$
$=\lambda_m^{\circ}(CH_3 COOK)+\frac{1}{2} \lambda_m^{\circ}(H_2 SO_4)-\frac{1}{2} \lambda_m^{\circ}(K_2 SO_4)$
$=z+\frac{x}{2}-\frac{y}{2}=z+(\frac{x-y}{2})$
