JEE Main202431 Jan 2024Evening ShiftChemistrySome Basic Concepts of ChemistryActual
A sample of CaCO 3 and MgCO 3 weighed 2 .21 g is ignited to constant weight of 1 .152 g . The composition of the mixture is: (Given molar mass in g mol − 1 , CaCO 3 : 100 , MgCO 3 : 84 )
Options
- A1 .187 g C a C O 3 + 1 .023 g M g C O 3
- B1 .023 g C a C O 3 + 1 .023 g M g C O 3
- C1 .187 g C a C O 3 + 1 .187 g M g C O 3
- D1 .023 g C a C O 3 + 1 .187 g M g C O 3
Correct answer
A. 1 .187 g C a C O 3 + 1 .023 g M g C O 3
Step-by-step solution
C a C O 3 s → Δ C a O s + C O 2 g M g C O 3 s → Δ M g O s + C O 2 g Let the weight of C a C O 3 be x g m ∴ weight of M g C O 3 = 2 .21 − x g m Moles of C a C O 3 decomposed = moles of C a O formed x 100 = moles of C a O formed ∴ weight of C a O formed = x 100 × 56 Moles of M g C O 3 decomposed = moles of M g O formed 2 .21 − x 84 = moles of M g O formed ∴ weight of M g O formed = 2 .21 − x 84 × 40 ⇒ 2 .21 − x 84 × 40 + x 100 × 56 = 1 .152 ∴ x = 1 .187 g = weight of C a C O 3 and weight of M g C O 3 = 1 .023 g