The equivalent weight of H_2O_2 is: | vedclass

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

Question

(A) The equivalent weight of a substance is calculated as the molar mass divided by the $n$-factor.For $H_2O_2$,the molar mass is $2 	imes 1 + 2 	im

Vedclass · Accepted Answer

$17$

Vedclass · Answer

(A) The equivalent weight of a substance is calculated as the molar mass divided by the $n$-factor.For $H_2O_2$,the molar mass is $2 	imes 1 + 2 	imes 16 = 34 \ g/mol$.In redox reactions,$H_2O_2$ acts as an oxidizing or reducing agent,typically involving a change in oxidation state of oxygen from $-1$ to $-2$ or $0$,resulting in an $n$-factor of $2$.Therefore,the equivalent weight = $\frac{	ext{Molar mass}}{n	ext{-factor}} = \frac{34}{2} = 17$.

The equivalent weight of $H_2O_2$ is:

Similar Questions

The $pH$ of an $H_2O_2$ solution is $6$. If some chlorine gas is passed through it,which of the following statements is correct?

$A$ commercial sample of hydrogen peroxide is labeled as $10$ volume. Its percentage concentration is approximately ............... $\%$.

The correct statements among the following are
$i$. Saline hydrides produce $H_2$ gas when reacted with water
$ii$. Presently $\sim 77 \%$ of the industrial dihydrogen is produced from coal
$iii$. Commercially marketed $H_2O_2$ contains $3 \% H_2O_2$

The volume strength of $1 \, M \, H_2O_2$ is : (Molar mass of $H_2O_2 = 34 \, g \, mol^{-1}$)

$KO_2$ reacts with water to form $A$,$B$ and $C$. $B$ forms $C$ when it reacts with iodine in basic medium. What are $B$ and $C$ respectively?

Vedclass Test Series

Exam Paper Generator

Online Exam Module