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Question

Calculate the molality of a 1 L soluttion of 93% H2SO4 (w/V) . The density of the solution is 1.84 g mL1

A
10.428 m
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B
9.428 m
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C
20.428 m
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D
18.428 m
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Solution

The correct option is A 10.428 m
93% w/V means that :
The solution is 1000 mL, containing 930 g of H2SO4.
The weight of the solution will be
=Density×Volume=(1.84×1000) g=1840 g

The weight of the solvent (H2O) :
Weight of solution - Weight of solute=(1840930) g=910 g
Molality=moles of H2SO4weight of H2O (g)×1000
=930/98910×1000=10.428 m.

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