Calculate the molality of $20 % (w / w) H_{2} O_{2} (a q)$ solution?

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Solution

$20 % (w / w)$ means $20 g$ of $H_{2} O_{2}$ is present in $100 g$ of the solution.
$Mass of solvent(water) = 100 g - 20 g = 80 g$

$Number of moles of H_{2} O_{2} = \frac{20 g}{molar mass of H_{2} O_{2}} = \frac{20 g}{34 g / m o l} = \frac{20}{34} m o l$
$Molality of solution = \frac{Number of moles of H_{2} O_{2}}{Mass of solvent in kg}$
$= \frac{20}{80 \times 34} \times 1000 = 7.35 m$

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