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Question

The molality of 15%(wt./vol.) solution of H2SO4 of density 1.1g/cm3 is approximately.

A
1.2
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B
1.4
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C
1.8
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D
1.6
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Solution

The correct option is D 1.6
The solution have 15% (mass/volume)

let mass of solute = 15g

volume of solution = 100 ml

we have, density = 1.1g/cm3

so, mass of solution = densityvolume=1.1×100=110g
mass of solvent(g)=11015g

molecular mass of H2SO4=98

so, moles of H2SO4 = given weightmolecular weight=1598=0.153moles

molality = moles of solutemass of solvent(g)×1000

so, molality =0.15395×1000=1.6


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