    Question

# Calculate the mass percent (w/w) of sulphuric acid in a solution prepared by dissolving 4 g of sulphur trioxide in a 100ml sulphuric acid solution containing 80 mass percent (w/w) of H2SO4 and having a density of 1.96 g/ml(molecular weight of H2SO4=98) Take reaction SO3+H2O→H2SO4.

A
80.8%
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B
84%
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C
41.65%
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D
None of these
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Solution

## The correct option is A 80.8%SO3+H2O→H2SO4100 ml sulphuric acid solution density = Massvolume1.96g/ml=Mass100Mass =1.96gAmount of water =196×20 =39.2gAmount of H2SO4=156.8gSO3+H2O→H2SO4Molar mass of SO3=32+16×3=80No. of mole = 4980g=0.05mol0.05 mol of sulphuric acid formedNow total amount of sulphuric acid =0.05×98+156.8=4.9+156.8=161.7total amount of water =39.2− total water consume =39.2−0.05×18 =39.2−0.9=38.3Mass percent (w/w) =161.7161.7+383=161.7200×100% =80.85% ≈80.8%Option A is correct.  Suggest Corrections  0      Similar questions  Related Videos   Percentage Composition and Molecular Formula
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