3.4 g of H_2O_2 is present in its 112 mL so | vedclass

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(D) First,calculate the number of moles of $H_2O_2$: $n = \frac{	ext{mass}}{	ext{molar mass}} = \frac{3.4 \ g}{34 \ g/mol} = 0.1 \ mol$. Next,calcul

Vedclass · Accepted Answer

$10$

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(D) First,calculate the number of moles of $H_2O_2$: $n = \frac{	ext{mass}}{	ext{molar mass}} = \frac{3.4 \ g}{34 \ g/mol} = 0.1 \ mol$. Next,calculate the molarity $(M)$ of the solution: $M = \frac{n}{V(L)} = \frac{0.1 \ mol}{112 	imes 10^{-3} \ L} = \frac{0.1}{0.112} \approx 0.8928 \ M$. The relationship between volume strength and molarity is: $	ext{Volume strength} = M 	imes 11.2$. Therefore,$	ext{Volume strength} = \frac{0.1}{0.112} 	imes 11.2 = \frac{0.1 	imes 11.2}{0.112} = \frac{1.12}{0.112} = 10 \ V$.

$3.4 \ g$ of $H_2O_2$ is present in its $112 \ mL$ solution. The volume strength of this solution is ....... $V$.

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