Mole ConceptMedium
Question
A 1.50 g sample of potassium bicarbonate having 80% purity is strongly heated. Assuming the impurity to be thermally stable, the loss in weight of the sample on heating is
Options
A.3.72 g
B.0.72 g
C.0.372 g
D.0.186 g
Solution
2KHCO3 → K2CO3 + CO2 ↑ + H2O↑
2 × 100 g (44 + 18) = 62 g
$\therefore 1.50 \times \frac{80}{100} = 1.20\text{ g} \frac{62}{200} \times 1.20 = 0.372\text{ g}$
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