A sample of CaCO 3 and MgCO 3 weighs 2.21\, g . The mixture is ignited to constant weight and the residue obtained is 1.152\, g . The composition of the mixture is: (Given molar masses (in g\,mol -1 ): CaCO 3 = 100 , MgCO 3 = 84 )

1.187\, g \;\ CaCO 3 + 1.023\, g \;\ MgCO 3

1.023\, g \;\ CaCO 3 + 1.023\, g \;\ MgCO 3

1.187\, g \;\ CaCO 3 + 1.187\, g \;\ MgCO 3

1.023\, g \;\ CaCO 3 + 1.187\, g \;\ MgCO 3

Chemistry · Some Basic Concepts of Chemistry (Mole Concept) + Stoichiometry

JEE Main 2024 — 31 January, Shift 2 — Question 59

A sample of $\mathrm{CaCO_3}$ and $\mathrm{MgCO_3}$ weighs $2.21\,\mathrm{g}$ . The mixture is ignited to constant weight and the residue obtained is $1.152\,\mathrm{g}$ . The composition of the mixture is:

(Given molar masses (in $\mathrm{g\,mol^{-1}}$ ): $\mathrm{CaCO_3} = 100$ , $\mathrm{MgCO_3} = 84$ )

Option A:
$1.187\,\mathrm{g}\;\ \mathrm{CaCO_3} + 1.023\,\mathrm{g}\;\ \mathrm{MgCO_3}$
Correct
Option B:
$1.023\,\mathrm{g}\;\ \mathrm{CaCO_3} + 1.023\,\mathrm{g}\;\ \mathrm{MgCO_3}$
Option C:
$1.187\,\mathrm{g}\;\ \mathrm{CaCO_3} + 1.187\,\mathrm{g}\;\ \mathrm{MgCO_3}$
Option D:
$1.023\,\mathrm{g}\;\ \mathrm{CaCO_3} + 1.187\,\mathrm{g}\;\ \mathrm{MgCO_3}$

Answer: A

Step-by-step solution

On heating, we have

\mathrm{CaCO_3(s) \xrightarrow{\Delta} CaO(s) + CO_2(g)}

\mathrm{MgCO_3(s) \xrightarrow{\Delta} MgO(s) + CO_2(g)}

Let mass of $\mathrm{CaCO_3} = x\,\mathrm{g}$ Then mass of $\mathrm{MgCO_3} = (2.21 - x)\,\mathrm{g}$

From molar masses:

100\,\mathrm{g\;\ CaCO_3} \rightarrow 56\,\mathrm{g\;\ CaO}

84\,\mathrm{g\;\ MgCO_3} \rightarrow 40\,\mathrm{g\;\ MgO}

Mass of residue $= \frac{56}{100}x + \frac{40}{84}(2.21 - x) = 1.152$

0.56x + 0.476(2.21 - x) = 1.152

0.084x = 0.101

x = 1.187

Hence, $\mathrm{CaCO_3} = 1.187\,\mathrm{g}, \;\mathrm{MgCO_3} = 1.023\,\mathrm{g}$

Thus, the correct option is A.

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 31 January, Shift 2
Subject: Chemistry
Chapter: Some Basic Concepts of Chemistry (Mole Concept) + Stoichiometry
Topic: Percentage composition & Empirical Formula

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Step-by-step solution

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