Calculate the strength of 20 V H_2O_2 in g/L | vedclass

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

Question

(B) The strength of $H_2O_2$ in $g/L$ is related to its volume strength by the formula: $	ext{Strength} (g/L) = 	ext{Volume strength} 	imes 3.03$.

Vedclass · Accepted Answer

$60.71$

Vedclass · Answer

(B) The strength of $H_2O_2$ in $g/L$ is related to its volume strength by the formula: $	ext{Strength} (g/L) = 	ext{Volume strength} 	imes 3.03$. Given,volume strength = $20 \ V$. Therefore,$	ext{Strength} = 20 	imes 3.03 = 60.6 \ g/L$. Using the more precise molar mass of $H_2O_2$ $(34.014 \ g/mol)$ and the molar volume of gas at $STP$ $(22.4 \ L)$: $	ext{Strength} = \frac{	ext{Volume strength} 	imes 68}{22.4} = \frac{20 	imes 68}{22.4} \approx 60.71 \ g/L$.

Calculate the strength of $20 \ V$ $H_2O_2$ in $g/L$.

Similar Questions

$A$ : The $O-O$ bond length in $H_2O_2$ is smaller than that in $O_2F_2$.
$R$ : $H_2O_2$ is an ionic compound.

$H_2O_2$ is a hydride of

Which of the following compounds is a peroxide?

The last traces of water present in $H_2O_2$ can be removed by:

To an aqueous solution of $AgNO_3$,some $NaOH_{(aq)}$ is added until a brown precipitate is obtained. To this,$H_2O_2$ is added dropwise. The precipitate turns black with the evolution of $O_2$. The black precipitate is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module