Assertion : The normality of 0.3 M aqueous s | vedclass

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(C) The normality of a solution is given by the formula: $	ext{Normality} = 	ext{Molarity} 	imes 	ext{n-factor}$.For $H_3PO_3$ (phosphorous acid),

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(C) The normality of a solution is given by the formula: $	ext{Normality} = 	ext{Molarity} 	imes 	ext{n-factor}$.For $H_3PO_3$ (phosphorous acid),the structure contains two $P-OH$ bonds,making it a dibasic acid. Thus,its n-factor is $2$.Normality $= 0.3 \ M 	imes 2 = 0.6 \ N$. Therefore,the Assertion is correct.The equivalent weight of an acid is defined as $\frac{	ext{Molecular weight}}{	ext{Basicity}}$.Since the basicity of $H_3PO_3$ is $2$,its equivalent weight is $\frac{	ext{Molecular weight}}{2}$.The Reason states the denominator is $3$,which is incorrect. Therefore,the Reason is incorrect.

Assertion : The normality of $0.3 \ M$ aqueous solution of $H_3PO_3$ is equal to $0.6 \ N$.
Reason : Equivalent weight of $H_3PO_3 = \frac{\text{Molecular weight of } H_3PO_3}{3}$

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Assertion : The normality of $0.3 \ M$ aqueous solution of $H_3PO_3$ is equal to $0.6 \ N$.Reason : Equivalent weight of $H_3PO_3 = \frac{\text{Molecular weight of } H_3PO_3}{3}$