    Question

# What is the molarity and normality of a 13% solution (by weight) of sulphuric acid with a density of 1.02 g/mL and also calculate the volume (in mL) that is required to prepare 1.5 N of this acid to make it upto 100ml?

A
Molarity=1.35, Normality=2.70, Volume=180 mL
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B
Molarity=2, Normality=4, Volume=200 mL
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C
Molarity=1.5, Normality=3, Volume=150 mL
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D
None of these
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Solution

## The correct option is A Molarity=1.35, Normality=2.70, Volume=180 mLMolarity=percentage by weight×10×dMw=13×1.02×1098=1.35 M13% solution by weight means 13 g of solute is dissolved in 87 g of solvent.m=W2×1000Mw×W1=1398×187×1000=1.52 mNormality=Molarity×2 ('n' factor for H2SO4)Normality=1.35×2=2.70For dilution,N1V1=N2V2or 100×2.70=1.5×V2or V2=180 mL  Suggest Corrections  0      Similar questions  Related Videos   Inequalities
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