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Question

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

A
10.4 m
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B
5.1 m
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C
12.7 m
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D
9.3 m
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Solution

The correct option is A 10.4 m
93% (w/v) means 93 g of the solute is present in 100 mL of the solution.
So, 1000 mL of the solution will contain 930 g of the solute.
Density of solution=weight of solutionvoulme of the solution
Weight of the solution = volume×density=1000×1.84=1840 g
Weight of the solvent
= weight of solution - weight of solute
Weight of the solvent
= 1840 - 930 = 910 g
Number of moles of H2SO4=given massmolar mass=93098=9.49 mol
Molality = moles of solute mass of solvent in g×1000
Molality = 9.49910×1000 = 10.4 m

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