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Molality of 100 grams of aqueous solution containing 20 grams of caustic soda is

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5 m

7.2 m

6.25 m

4.2 m

answer is C.

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Detailed Solution

$W_{solute} = 20 g; GMW of solute = 40 g; W_{solution} = 100 g; W_{solvent} = (100 - 20) g; m = \frac{20}{40} \times \frac{1000}{80} = 6.25 m$

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