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JEE CHEMISTRY MCQ: SOME BASIC CONCEPTS OF CHEMISTRY

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Analyzing +4 | -1
Assertion (A) : 20g20 \mathrm{g} of 50%50 \% pure CaCO3\mathrm{CaCO}_3 on heating gave 2.24 L CO2\mathrm{CO}_2 at STP and 15.6 g15.6 \mathrm{~g} non-volatile residue.

CaCO3⟶ΔCaO+CO2\mathrm{CaCO}_3 \stackrel{\Delta}{\longrightarrow} \mathrm{CaO}+\mathrm{CO}_2
Reason (R): Pure CaCO3\mathrm{CaCO}_3 is 10 g10 \mathrm{~g} which gave 4.4 g4.4 \mathrm{~g} CO2(=0.1 mol=2.24 L\mathrm{CO}_2(=0.1 \mathrm{~mol}=2.24 \mathrm{~L} at STP )) and 15.6 g15.6 \mathrm{~g} residue is left.
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Correct option: A
Both A\mathbf{A} and R\mathbf{R} are true and R\mathbf{R} is the correct explanation of A\mathbf{A}.
The correct answer is A because both the assertion and the reason are true, and the reason correctly explains why the given amount of impure CaCO3 produces 2.24 L of CO2 at STP.

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