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Question

0.1 N solution of a dibasic acid can be prepared by dissolving 0.45 g of the acid in water and diluting to 100 mL. The molecular mass of the acid is:

A
45
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B
90
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C
135
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D
180
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Solution

The correct option is D 90
Normality =numberofgramequivalentvolumeofsolutioninlitre
numberofgramequivalentvolumeofsolutioninlitreNormality=massofsoluteequivalentmass×1000Vol(ml)0.1=0.45eq.mass×1000100eq.mass=0.450.1×10eq.mass=45Asitisadibasicacid,somolecularmass=2×equivalentmass=2×45Molecularmass=90

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