Calculate the percentage by mass of a H_2O_2 | vedclass

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Question

(B) The volume strength of $H_2O_2$ is related to its molarity $(M)$ and percentage by mass $(w/w \%)$.
The relation is: $	ext{Volume strength} = 11

Vedclass · Accepted Answer

$20.40 \%$ by mass

Vedclass · Answer

(B) The volume strength of $H_2O_2$ is related to its molarity $(M)$ and percentage by mass $(w/w \%)$.
The relation is: $	ext{Volume strength} = 11.2 	imes M$.
Given volume strength = $67.2$.
So,$M = \frac{67.2}{11.2} = 6 \ M$.
The relation between molarity $(M)$ and percentage by mass $(w/w \%)$ is: $M = \frac{w/w \% 	imes d 	imes 10}{M_w}$,where $d$ is density and $M_w$ is molar mass of $H_2O_2$ $(34 \ g/mol)$.
Assuming density $d \approx 1 \ g/mL$ for dilute solutions,$6 = \frac{w/w \% 	imes 1 	imes 10}{34}$.
$w/w \% = \frac{6 	imes 34}{10} = 20.40 \%$.

Calculate the percentage by mass of a $H_2O_2$ solution that is $67.2$ volume strength.

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