20.0 g 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?
$(At . wt . of Mg = 24)$

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answer is C.

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Detailed Solution

$\begin{matrix} \underset{84 g}{{MgCO}_{3 (s)}} \overset{Δ}{⟶} \underset{40 g}{{MgO}_{(s)}} + {CO}_{2 (g)} \end{matrix}$

$84 g of {MgCO}_{3} \equiv 40 g of MgO$

$∴ 20 g of {MgCO}_{3} \equiv \frac{40}{84} \times 20 = 9.52 g of MgO$

$Actual yield = 8 g of MgO$

$∴ % purity = \frac{8}{9.52} \times 100 = 84 %$

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