Calculate the weight of CaCO3 required to produce a 0.5N of 100mL solution.
A
2.5g
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B
5g
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C
7.5g
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D
10g
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Solution
The correct option is A2.5g Let the weight of CaCO3 be xg. Molecular weight of CaCO3=40+12+(3×16)=100gmol−1 Volume=1001000L=0.1Lnf=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=2.5g ∴ weight of CaCO3 required is 2.5g.