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Question

Calculate the molality of a 1 L solution of 68% H2SO4(wv), if the density of the solution is given as 1.80 g mL1.

A
6.8 m
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B
6.2 m
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C
6.94 m
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D
10.8 m
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Solution

The correct option is B 6.2 m
68% (wv)H2SO4=68 g H2SO4 in 100 mL solution.
i.e., 680 g H2SO4 in 1000 mL solution.
Mass of solution = volume × density
=1000 mL×1.80g/mL
=1800 g
Mass of solvent = 1800680=1120 g
Again, Molar mass of H2SO4=2×1+32+16×4=98 g
Number of moles of solute,
MassMolar mass=68098=6.94 mol
Molality:
=No. of moles of soluteMass of solvent in kg=(6.94×1000)1120=6.2 m

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