What is 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 40 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 $H_{2} O_{2} = \frac{25.5 g}{m o l a r m a s s o f H_{2} O_{2}} = \frac{25.5 g}{34 g / m o l} = \frac{25.5}{34} m o l$
$M o l a l i t y o f s o l u t i o n = \frac{n u m b e r o f m o l e s o f H_{2} O_{2}}{m a s s o f s o l v e n t i n k g}$
$= \frac{25.5}{74.5 \times 34} \times 1000 = 10.06 m$

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