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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 ........... $g$.

Similar Questions

$H_2O_2$ is used as:

The environmentally friendly chemical used nowadays for bleaching paper in the presence of a suitable catalyst is

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

Which one of the following reactions represents the oxidising property of $H_2O_2$?

The reaction $H_2S + H_2O_2 \to S + 2H_2O$ manifests

Vedclass Test Series

Exam Paper Generator

Online Exam Module