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(A) The strength of $H_2O_2$ is given as $8.32\% \ W/V$,which means $8.32 \ g$ of $H_2O_2$ is present in $100 \ mL$ of solution.$1$. Calculation of Vo

Vedclass · Accepted Answer

$27.4, 2.447$

Vedclass · Answer

(A) The strength of $H_2O_2$ is given as $8.32\% \ W/V$,which means $8.32 \ g$ of $H_2O_2$ is present in $100 \ mL$ of solution.$1$. Calculation of Volume Strength:Volume strength of $H_2O_2 = 	ext{Strength in } g/L 	imes \frac{11.2}{34}$.Strength in $g/L = 8.32 	imes 10 = 83.2 \ g/L$.Volume strength $= 83.2 	imes \frac{11.2}{34} \approx 27.4 \ V$.$2$. Calculation of Molar Concentration:Molarity $(M) = \frac{	ext{Strength in } g/L}{	ext{Molar mass of } H_2O_2}$.Molar mass of $H_2O_2 = (2 	imes 1) + (2 	imes 16) = 34 \ g/mol$.$M = \frac{83.2}{34} \approx 2.447 \ M$.Thus,the volume strength is $27.4 \ V$ and molarity is $2.447 \ M$.

The strength of $H_2O_2$ sample available from market is $8.32\% \ W/V$. The volume and molar concentration of this sample are ......... and ........ respectively.

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