Calculate the molality of $25.5 %$ (w/w) $H_{2} O_{2}$ aqueous solution? (Give your answer upto two decimal places.)

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Solution

$25.5 %$ (w/w) means $25.5 g$ of $H_{2} O_{2}$ is present in $100 g$ of the solution.
Mass of solvent $= 100 g - 25.5 g = 74.5 g$
Number of moles of $H_{2} O_{2} = \frac{25.5 g}{molar mass of H_{2} O_{2}} = \frac{25.5 g}{34 g / m o l} = \frac{25.5}{34} m o l$
$Molality of solution = \frac{number of moles of H_{2} O_{2}}{mass of solvent in kg}$
$= \frac{25.5}{74.5 \times 34} \times 1000 = 10.06 m$

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