Question

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Chapter: Volumetric Analysis(Titration)(Acidimetry and Alkalimetry)

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Answer 1

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Given,


D=1.25g/ml

Malvity=3

Mass of sodium = 125×1000ml

=1250g

No.of moles = 3

Mass of No2 So2 = 3(2×33+64+3×16)

=3×(16+64+48)

=3×158

=174

Percentage mass = 474/1256×100

=37.92

Mass of slovent = 1250/7768-174

Malvity of Na2 = 3/776/1000

=3.86



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Answer 2

Answer:

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