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Molality

What is molal...

Question

What is molality of 30% (w/w) $H_{2} O_{2}$ aqueous solution?

A

8.82 m

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B

12.60 m

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C

15.60 m

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D

10.67 m

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Solution

The correct option is B 12.60 m
30% (w/w) means 30 g of $H_{2} O_{2}$ is present in 100 g of the solution.
Mass of solvent = 100 g - 30 g = 70 g
Number of moles of $H_{2} O_{2} = \frac{30 g}{m o l a r m a s s o f H_{2} O_{2}} = \frac{30 g}{34 g / m o l} = \frac{30}{34} m o l$
$M o l a l i t y o f s o l u t i o n = \frac{n u m b e r o f m o l e s o f H_{2} O_{2}}{m a s s o f s o l v e n t i n k g}$
$= \frac{30}{70 \times 34} \times 1000 = 12.60 m$

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