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Question

Calculate normality of the solution if 5.3 g of solid Na2CO3 is dissolved to make 250 mL of the solution.

A
0.1 N
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B
0.4 N
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C
0.75 N
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D
1 N
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Solution

The correct option is B 0.4 N
Molecular weight of Na2CO3=(23×2)+12+(3×16)=106 g mol1Volume=2501000 L=0.25 Lnf=2

Since,
Normality=Number of g-equivalents of soluteVolume of solution (L)Number of g-equivalents of solute=Weight of soluteEquivalent weight of soluteEquivalent weight of solute=Molecular Mass of soluten-factor

Putting the values we get,

Normality=Weight of solute×n-factorMolecular Mass of solute×Volume of solution (L)Normality=5.3 g×2106 g mol1×0.25 LNormality=0.4 N

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