A given sample of sea water is 2.925% (wt/wt). The molality of the solution is approximately:

5.2

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0.52

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0.94

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1.02

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Solution

The correct option is B 0.52
$\frac{w t}{w t}$ percentage= $\frac{W e i g h t o f s o l u t e}{W e i g h t o f s o l u t i o n} \times 100$
$⟹ \frac{2.925}{100} = \frac{W e i g h t o f s o l u t e (W_{s})}{W_{s} + W e i g h t o f s o l v e n t}$
Considering one mole of water is taken, as it is sea water.
$⟹ 0.0295 = \frac{W_{s}}{W_{s} + 18}$
$⟹ (0.0295) (W_{s} + 18) = W_{s}$
$⟹ W_{s} (1 - 0.0295) = 18 (0.0295)$
$⟹ W_{s} (1 - 0.0295) = 0.531$
$⟹ W_{s} = 0.536$
Molality= $\frac{W_{s}}{G M W o f s o l u t e} \times \frac{1000}{W e i g h t o f s o l v e n t (g r a m s)} = \frac{0.536}{58.5} \times \frac{1000}{18} = \frac{536}{1053} = 0.52$

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