Molarity of perhydrol is

Perhydrol is 30.35%  $\left(\frac{\mathrm{w}}{\mathrm{v}}\right){\mathrm{H}}_2{\mathrm{O}}_2\mathrm{solution}$

$\begin{array}{l}\mathrm{M}=\frac{10\times \left(\frac{\mathrm{w}}{\mathrm{v}}\mathrm{\%}\right)}{\mathrm{GMW}\mathrm{of}\mathrm{solute}}\\ \mathrm{M}=\frac{10\times 30.35}{34}=\frac{303.5}{34}=8.9\mathrm{M}\end{array}$