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Question

6.2 g of a sample containing Na2CO3,NaHCO3 and non-volatile inert impurity on gentle heating loses 5% of its weight due to reaction: 2NaHCO3Na2CO3+H2O+CO2. Residue is dissolved in water and formed 100mL solution and its 10mL portion requires 7.5mL of 0.2M aqueous solution of BaCl2 for complete precipitation of carbonates. Determine weight (in g) of Na2CO3 in the original sample.

A
1.59
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B
1.06
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C
0.53
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D
None of these
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Solution

The correct option is B 1.06
Loss in weight due to heating=6.2×5100=0.31 g.
2NaHCO3Na2CO3+CO2+H2O
2×84g44+18g loss
1 mole NaHCO3622=31g loss due to heating
0.31g loss from 0.01 mol of NaHCO3.
Moles of Na2CO3 produced =0.012=0.005.
Total moles of carbonate reacted with BaCl2=(7.5×0.2×10010)×103=0.015
Moles of carbonates in original sample=0.0150.005=0.001
Mass of Na2CO3 in original sample=0.01×106=1.06g

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