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Question

20 g of a sample of Ba(OH)2 is dissolved in 10 mL of 0.5 N HCl solution. The excess of HCl was titrated with 0.2 N NaOH. The volume of NaOH required was 10 mL. Calculate the percentage of Ba(OH)2 in the sample.
(Given molar mass of Ba is 137 g mol1)

A
12.8 %
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B
1.28 %
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C
50 %
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D
33.33 %
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Solution

The correct option is B 1.28 %
Milliequivalents of HCl present initially=10×0.5=5
Milliequivalents of NaOH consumed = Milliequivalents of HCl in excess =10×0.2=2
Milliequivalents of HCl consumed=Milliequivalents of Ba(OH)2=52=3
Equivalents of Ba(OH)2=3/1000=3×103
Molecular Mass of Ba(OH)2=137+(16+1)×2=171 g mol1
Mass of Ba(OH)2=3×103×1712=0.2565 g
% Ba(OH)2=0.256520×100=1.28 %

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