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Question

Calculate the weight of CaCO3 required to produce a 0.5 N of 100 mL solution.

A
2.5 g
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B
5 g
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C
7.5 g
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D
10 g
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Solution

The correct option is A 2.5 g
Let the weight of CaCO3 be x g.
Molecular weight of CaCO3=40+12+(3×16)=100 g mol1
Volume=1001000 L=0.1 Lnf=2
Normality=Number of g-equivalents of soluteVolume of solution (L)

Normality=Weight of solute×n-factorMolecular Mass of solute×Volume of solution (L)0.5=x×2100×0.1x=2.5 g
weight of CaCO3 required is 2.5 g.

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