40 gm of a magnesium carbonate sample decomposes on heating to give carbon dioxide and 16 gm magnesium oxide. What will be the percentage purity of magnesium carbonate in the sample.

60%

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84%

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75%

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96%

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Solution

The correct option is B 84%
$M g C O_{3} \to M g O + C O_{2}$

1 mole of $M g O$ is formed from 1 mole of $M g C O_{3}$

40 g of $M g O$ is formed from 84 g of $M g C O_{3}$

Hence mass of $M g C O_{3}$ present in the sample $= \frac{16}{40} \times 84 = \frac{2}{5} \times 84$
$percentage of purity = \frac{2 \times 84}{40 \times 5} \times 100 = 84 %$

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$20.0 g$ of a magnesium carbonate sample decomposes on heating to give carbon dioxide and $8.0 g$ of magnesium oxide. What will be the percentage purity of magnesium carbonate in the sample? (At. Wt. $M g = 24$ )

Q. 20.0g of a magnesium carbonate sample decomposes on heating to give carbon dioxide and 8.0 g magnesium oxide. What will be the percentage purity of magnesium carbonate in the sample?
(At. Wt. Mg = 24)

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40 gm of a magnesium carbonate sample decomposes on heating to give carbon dioxide and 16 gm magnesium oxide. What will be the percentage purity of magnesium carbonate in the sample.

The correct option is B 84%MgCO3→MgO+CO2 1 mole of MgO is formed from 1 mole of MgCO3 40 g of MgO is formed from 84 g of MgCO3 Hence mass of MgCO3 present in the sample =1640×84=25×84 percentage of purity=2×8440×5×100=84%