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Q)

$4g$ of impure $CaCO_3$ on treatment with excess HCl produces $0.88\; g\; CO_2$. What is the percent purity of $CaCO_3$ sample?

$(a)\;75 \\ (b)\;25 \\(c)\;90 \\(d)\;50 $

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A)
Solution :
$CaCO_3+2HCl \to CaCl_2+CO_2+H_2O$
here, In 44 g of $\to$ 100 g of
So, for $0.88\;g\;of\;CO_2\to(100/44)*0.88\;g=2\;g$
% of purity $= \large\frac{2}{4}*100=50\%$
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