For preparing 0.05 N H_2SO_4 solution in on | vedclass

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(D) The normality $(N)$ of a solution is given by the formula: $N = \frac{	ext{Mass of solute (g)}}{	ext{Equivalent weight} 	imes 	ext{Volume of s

Vedclass · Accepted Answer

$2.45$

Vedclass · Answer

(D) The normality $(N)$ of a solution is given by the formula: $N = \frac{	ext{Mass of solute (g)}}{	ext{Equivalent weight} 	imes 	ext{Volume of solution (L)}}$.For $H_2SO_4$,the molar mass is $98 \ g/mol$. The basicity of $H_2SO_4$ is $2$,so its equivalent weight is $\frac{98}{2} = 49 \ g/eq$.Given: $N = 0.05 \ N$,Volume = $1 \ L$,Equivalent weight = $49 \ g/eq$.Substituting these values: $0.05 = \frac{	ext{Mass}}{49 	imes 1}$.Mass = $0.05 	imes 49 = 2.45 \ g$.

For preparing $0.05 \ N \ H_2SO_4$ solution in one litre,we need $H_2SO_4$. (in $g$)

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