JEE Main · 2024 · Shift-IIhardMOLE-162

A sample of CaCO3 and MgCO3 weighed 2.21\,g is ignited to constant weight of 1.152\,g. The composition of the mixture…

Some Basic Concepts (Mole Concept) · Class 11 · JEE Main Previous Year Question

Question

A sample of $\ce{CaCO3}$ and $\ce{MgCO3}$ weighed $2.21\,\text{g}$ is ignited to constant weight of $1.152\,\text{g}$ . The composition of the mixture is: ( $M_r$ : $\ce{CaCO3}=100$ , $\ce{MgCO3}=84$ )

Options

a
$1.187\,\text{g}\;\ce{CaCO3} + 1.023\,\text{g}\;\ce{MgCO3}$
✓
b
$1.023\,\text{g}\;\ce{CaCO3} + 1.023\,\text{g}\;\ce{MgCO3}$
c
$1.187\,\text{g}\;\ce{CaCO3} + 1.187\,\text{g}\;\ce{MgCO3}$
d
$1.023\,\text{g}\;\ce{CaCO3} + 1.187\,\text{g}\;\ce{MgCO3}$

Correct Answera

$1.187\,\text{g}\;\ce{CaCO3} + 1.023\,\text{g}\;\ce{MgCO3}$

Detailed Solution

🧠 On heating, carbonate → oxide + $\ce{CO2}$ $\ce{CaCO3} \to \ce{CaO} + \ce{CO2}$ keeps $56\,\text{g}$ out of $100$ . $\ce{MgCO3} \to \ce{MgO} + \ce{CO2}$ keeps $40\,\text{g}$ out of $84$ .

Set up two equations: total mass and total residue.

🗺️ Solve the system Let $x =$ mass of $\ce{CaCO3}$ , $y =$ mass of $\ce{MgCO3}$ . $x + y = 2.21.$ $\frac{56}{100}x + \frac{40}{84}y = 1.152.$ i.e. $0.56 x + 0.476 y = 1.152$ . Substitute $y = 2.21 - x$ : $0.56 x + 0.476 (2.21 - x) = 1.152.$ $0.084 x = 0.152 - 0 = 0.100 \Rightarrow x \approx 1.187\,\text{g}.$ Then $y = 2.21 - 1.187 = 1.023\,\text{g}$ .

So the mixture is $1.187\,\text{g}$ $\ce{CaCO3} + 1.023\,\text{g}$ $\ce{MgCO3}$ .

$\boxed{\text{Answer: (a)}}$

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