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Some Basic Concepts of Chemistry question

2024 · 31 Jan · Shift 2 · Q20
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Some Basic Concepts of Chemistry question

2024 · 31 Jan · Shift 2 · Q20

JEE MainChemistrySome Basic Concepts of ChemistryMCQ+4 / −1
A sample of CaCO3\mathrm{CaCO}_3CaCO3​ and MgCO3\mathrm{MgCO}_3MgCO3​ weighed 2.21 g2.21 \mathrm{~g}2.21 g is ignited to constant weight of 1.152 g1.152 \mathrm{~g}1.152 g. The composition of mixture is : (Given molar mass in g mol−1CaCO3:100,MgCO3:84\mathrm{g} \mathrm{~mol}^{-1} \mathrm{CaCO}_3: 100, \mathrm{MgCO}_3: 84g mol−1CaCO3​:100,MgCO3​:84)
  1. A
    1.187 g CaCO3+1.187 g MgCO31.187 \mathrm{~g} \mathrm{~CaCO}_3+1.187 \mathrm{~g} \mathrm{~MgCO}_31.187 g CaCO3​+1.187 g MgCO3​
  2. B
    1.187 g CaCO3+1.023 g MgCO31.187 \mathrm{~g} \mathrm{~CaCO}_3+1.023 \mathrm{~g} \mathrm{~MgCO}_31.187 g CaCO3​+1.023 g MgCO3​
  3. C
    1.023 g CaCO3+1.187 g MgCO31.023 \mathrm{~g} \mathrm{~CaCO}_3+1.187 \mathrm{~g} \mathrm{~MgCO}_31.023 g CaCO3​+1.187 g MgCO3​
  4. D
    1.023 g CaCO3+1.023 g MgCO31.023 \mathrm{~g} \mathrm{~CaCO}_3+1.023 \mathrm{~g} \mathrm{~MgCO}_31.023 g CaCO3​+1.023 g MgCO3​
View written solutionFree

Correct answer: B

  1. Write the decomposition reactions on ignition

On heating,

CaCO3→CaO+CO2\mathrm{CaCO_3 \rightarrow CaO + CO_2}CaCO3​→CaO+CO2​ MgCO3→MgO+CO2\mathrm{MgCO_3 \rightarrow MgO + CO_2}MgCO3​→MgO+CO2​

After ignition, only the oxides remain.

  1. Mass loss on ignition

Initial mass of mixture:

2.21 g2.21\,\text{g}2.21g

Final constant mass after ignition:

1.152 g1.152\,\text{g}1.152g

So mass of CO2\mathrm{CO_2}CO2​ lost is

2.21−1.152=1.058 g2.21-1.152=1.058\,\text{g}2.21−1.152=1.058g
  1. Let masses of the carbonates be
x=mass of CaCO3x = \text{mass of } \mathrm{CaCO_3}x=mass of CaCO3​ y=mass of MgCO3y = \text{mass of } \mathrm{MgCO_3}y=mass of MgCO3​

Then,

x+y=2.21x+y=2.21x+y=2.21
  1. Relate carbonate mass to } \mathrm{CO_2} \text{ evolved}

From molar masses:

  • 100 g100\,\text{g}100g of CaCO3\mathrm{CaCO_3}CaCO3​ gives 44 g44\,\text{g}44g of CO2\mathrm{CO_2}CO2​
  • 84 g84\,\text{g}84g of MgCO3\mathrm{MgCO_3}MgCO3​ gives 44 g44\,\text{g}44g of CO2\mathrm{CO_2}CO2​

Therefore,

44100x+4484y=1.058\frac{44}{100}x + \frac{44}{84}y = 1.05810044​x+8444​y=1.058

So,

0.44x+1121y=1.0580.44x + \frac{11}{21}y = 1.0580.44x+2111​y=1.058

Using x=2.21−yx=2.21-yx=2.21−y,

0.44(2.21−y)+1121y=1.0580.44(2.21-y)+\frac{11}{21}y=1.0580.44(2.21−y)+2111​y=1.058 0.9724−0.44y+0.523809y=1.0580.9724 - 0.44y + 0.523809y = 1.0580.9724−0.44y+0.523809y=1.058 0.083809y=1.058−0.9724=0.08560.083809y = 1.058 - 0.9724 = 0.08560.083809y=1.058−0.9724=0.0856 y≈0.08560.083809≈1.02 gy \approx \frac{0.0856}{0.083809} \approx 1.02\,\text{g}y≈0.0838090.0856​≈1.02g

Hence,

y≈1.023 gy \approx 1.023\,\text{g}y≈1.023g

and

x=2.21−1.023=1.187 gx = 2.21 - 1.023 = 1.187\,\text{g}x=2.21−1.023=1.187g
  1. Composition of the mixture
1.187 g of CaCO3 and 1.023 g of MgCO3\boxed{1.187\,\text{g of } \mathrm{CaCO_3} \text{ and } 1.023\,\text{g of } \mathrm{MgCO_3}}1.187g of CaCO3​ and 1.023g of MgCO3​​
  1. Check with options

This matches Option B.

  1. Comparison with stored correct answer

Stored correct answer: B

My derived answer: B

So, the answer agrees with the stored correct answer.

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