6.8 , g of H_2O_2 is dissolved in 0.2 , L of | vedclass

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(B) $1$. Calculate the molarity $(M)$ of the $H_2O_2$ solution:$M = \frac{	ext{mass of } H_2O_2}{	ext{molar mass of } H_2O_2 	imes 	ext{volume of

Vedclass · Accepted Answer

$5.6$

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(B) $1$. Calculate the molarity $(M)$ of the $H_2O_2$ solution:$M = \frac{	ext{mass of } H_2O_2}{	ext{molar mass of } H_2O_2 	imes 	ext{volume of solution in } L}$$M = \frac{6.8 \, g}{34 \, g/mol 	imes 0.2 \, L} = \frac{6.8}{6.8} = 1 \, M$$2$. Use the relation between volume strength and molarity:$	ext{Volume strength} = M 	imes 11.2$$	ext{Volume strength} = 1 	imes 11.2 = 11.2 \, V$$3$. Re-evaluating the options provided,there seems to be a discrepancy. However,if we consider the calculation $11.2 \, V$,none of the options match exactly. Given the standard context of such problems,if the volume was $0.4 \, L$,the answer would be $5.6 \, V$. Based on the provided values,the calculated result is $11.2 \, V$.

$6.8 \, g$ of $H_2O_2$ is dissolved in $0.2 \, L$ of solution. Then the solution will be labelled as ............. $V$ volume strength.

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