The % left(frac{W}{V}right) of H_2O_2 in 1.5 | vedclass

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(C) For $H_2O_2$,the relationship between Normality $(N)$ and Molarity $(M)$ is given by: $N = n 	imes M$,where $n$-factor for $H_2O_2$ is $2$. $M =

Vedclass · Accepted Answer

$2.55$

Vedclass · Answer

(C) For $H_2O_2$,the relationship between Normality $(N)$ and Molarity $(M)$ is given by: $N = n 	imes M$,where $n$-factor for $H_2O_2$ is $2$. $M = \frac{N}{2} = \frac{1.5}{2} = 0.75 \ M$. $\% \left(\frac{W}{V}ight)$ is defined as the mass of solute in grams present in $100 \ mL$ of solution. Mass of $H_2O_2$ in $1 \ L$ $(1000 \ mL)$ $= M 	imes 	ext{Molar Mass} = 0.75 	imes 34 = 25.5 \ g/L$. Therefore,in $100 \ mL$,the mass is $\frac{25.5}{1000} 	imes 100 = 2.55 \ g$. Thus,the $\% \left(\frac{W}{V}ight)$ is $2.55$.

The $\% \left(\frac{W}{V}\right)$ of $H_2O_2$ in $1.5 \ N$ solution is

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