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Molarity of perhydrol is

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10 M

4.45 M

17.8 M

8.9 M

answer is D.

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Detailed Solution

Perhydrol is 30.35% $(\frac{w}{v}) H_{2} O_{2} solution$

$\begin{array}{l} M = \frac{10 \times (\frac{w}{v} %)}{GMW of solute} \\ M = \frac{10 \times 30 .35}{34} = \frac{303 .5}{34} = 8 .9 M \end{array}$

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