The amount of H_2O_2 present in 1 L of 1.5 | vedclass

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(B) Strength of the solution is given by the formula: $	ext{Strength} = 	ext{Normality} 	imes 	ext{Equivalent mass}$.The equivalent mass of $H_2O_

Vedclass · Accepted Answer

$25.5$

Vedclass · Answer

(B) Strength of the solution is given by the formula: $	ext{Strength} = 	ext{Normality} 	imes 	ext{Equivalent mass}$.The equivalent mass of $H_2O_2$ is $\frac{	ext{Molar mass}}{n	ext{-factor}} = \frac{34}{2} = 17 \ g \ eq^{-1}$.Given,$	ext{Normality} = 1.5 \ N$ and $	ext{Volume} = 1 \ L$.$	ext{Strength} = 1.5 	imes 17 = 25.5 \ g \ L^{-1}$.Since the volume is $1 \ L$,the amount of $H_2O_2$ present is $25.5 \ g$.

The amount of $H_2O_2$ present in $1 \ L$ of $1.5 \ N$ $H_2O_2$ solution is ........... $g$.

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