The conductance of a 0.0015 M aqueous soluti | vedclass

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(C) The cell constant $G^* = \frac{\ell}{a} = \frac{120 \ cm}{1 \ cm^2} = 120 \ cm^{-1}$.The conductivity $\kappa = G 	imes G^* = 5 	imes 10^{-7} \

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(C) The cell constant $G^* = \frac{\ell}{a} = \frac{120 \ cm}{1 \ cm^2} = 120 \ cm^{-1}$.The conductivity $\kappa = G 	imes G^* = 5 	imes 10^{-7} \ S 	imes 120 \ cm^{-1} = 6 	imes 10^{-5} \ S \ cm^{-1}$.The molar conductivity at concentration $C$ is $\Lambda_{m}^{c} = \frac{\kappa 	imes 1000}{C} = \frac{6 	imes 10^{-5} 	imes 1000}{0.0015} = 40 \ S \ cm^2 \ mol^{-1}$.For a weak acid,$pH = 4 \implies [H^+] = 10^{-4} \ M$. Since $[H^+] = C \alpha$,the degree of dissociation $\alpha = \frac{10^{-4}}{0.0015} = \frac{1}{15}$.Using the relation $\alpha = \frac{\Lambda_{m}^{c}}{\Lambda_{m}^0}$,we get $\Lambda_{m}^0 = \frac{\Lambda_{m}^{c}}{\alpha} = \frac{40}{1/15} = 600 \ S \ cm^2 \ mol^{-1}$.Given $\Lambda_{m}^0 = Z 	imes 10^2 \ S \ cm^2 \ mol^{-1}$,we have $Z 	imes 10^2 = 600$,which implies $Z = 6$.

Electrolyte	$\Lambda ^{\infty } (S\ cm^2\ mol^{-1})$
$KCl$	$149.9$
$KNO_3$	$145.0$
$HCl$	$426.2$
$NaOAc$	$91.0$
$NaCl$	$126.5$

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