The volume strength of 1.5 N H_2O_2 solutio | vedclass

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

Question

(C) Strength = $	ext{Normality} 	imes 	ext{Equivalent Weight (EW)}$ of $H_2O_2$$= 1.5 \ N 	imes 17 \ g \ eq^{-1} = 25.5 \ g \ L^{-1}$Decomposition

Vedclass · Accepted Answer

$8.4$

Vedclass · Answer

(C) Strength = $	ext{Normality} 	imes 	ext{Equivalent Weight (EW)}$ of $H_2O_2$$= 1.5 \ N 	imes 17 \ g \ eq^{-1} = 25.5 \ g \ L^{-1}$Decomposition reaction: $2H_2O_2 ightarrow 2H_2O + O_2$$2 	imes 34 \ g = 68 \ g$ of $H_2O_2$ produces $22.4 \ L$ of $O_2$ at $STP$.Volume strength = $\frac{11.2 	imes 	ext{Normality}}{2} = 5.6 	imes 	ext{Normality}$$= 5.6 	imes 1.5 = 8.4$Thus,the volume strength of $1.5 \ N \ H_2O_2$ solution is $8.4$.

$A$. Peroxide	$1$. $C_3O_2$
$B$. Superoxide	$2$. $PbO_2$
$C$. Dioxide	$3$. $KO_2$
$D$. Suboxide	$4$. $H_2O_2$

The volume strength of $1.5 \ N \ H_2O_2$ solution is

Similar Questions

Fenton’s reagent is

All the products formed in the oxidation of $NaBH_4$ by $I_2$ are

Give the reaction of $I_2$ with $H_2O_2$ in the presence of a basic medium.

Match List $I$ with List $II$ and select the correct answer using the codes given below the lists:
$A$. Peroxide $1$. $C_3O_2$
$B$. Superoxide $2$. $PbO_2$
$C$. Dioxide $3$. $KO_2$
$D$. Suboxide $4$. $H_2O_2$

Blackened oil painting can be restored into original form by the action of :

Vedclass Test Series

Exam Paper Generator

Online Exam Module