For an HCl solution at 25^o C,the equivalent | vedclass

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(C) The degree of dissociation $\alpha$ is given by the ratio of equivalent conductance at concentration $C$ to equivalent conductance at infinite dil

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$10$

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(C) The degree of dissociation $\alpha$ is given by the ratio of equivalent conductance at concentration $C$ to equivalent conductance at infinite dilution: $\alpha = \frac{\wedge_{eq}^c}{\wedge_{eq}^0}$.We know that $\wedge_{eq}^c = \frac{\kappa 	imes 1000}{N}$,where $\kappa$ is the conductivity and $N$ is the normality.Given: $\alpha = 0.90$,$\wedge_{eq}^0 = 425 \ \Omega^{-1} \ cm^2 \ eq^{-1}$,and $\kappa = 3.825 \ \Omega^{-1} \ cm^{-1}$.Substituting the values: $0.9 = \frac{3.825 	imes 1000}{N 	imes 425}$.$0.9 = \frac{3825}{N 	imes 425}$.$0.9 = \frac{9}{N}$.$N = \frac{9}{0.9} = 10 \ N$.

Ion and $\lambda ^o / (S \ cm^2 \ mol^{-1})$	Ion and $\lambda ^o / (S \ cm^2 \ mol^{-1})$
$H^{+} : 349.6$	$OH^{-} : 199.1$
$Na^{+} : 50.1$	$Cl^{-} : 76.3$
$K^{+} : 73.5$	$Br^{-} : 78.1$
$Ca^{2+} : 119.0$	$CH_3COO^{-} : 40.9$
$Mg^{2+} : 106.0$	$SO_4^{2-} : 160.0$

For an $HCl$ solution at $25^o C$,the equivalent conductance at infinite dilution is $425 \ \Omega^{-1} \ cm^2 \ eq^{-1}$. The electrical conductivity of the $HCl$ solution is $3.825 \ \Omega^{-1} \ cm^{-1}$. If the degree of dissociation is $90\%$,what is the normality of the solution in $N$?

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