Chemistry - Mole Concept Question with Solution | TestHub

ChemistryMole ConceptMiscellaneous/MixedMedium2 minPYQ_2024
ChemistryMediumassertion reason

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

(Given molar mass in g mol−1, CaCO3:100, MgCO3:84)

Options:

Answer:
A
Solution:

CaCO3s→ΔCaOs+CO2g

MgCO3s→ΔMgOs+CO2g

Let the weight of CaCO3 be x gm

∴ weight of MgCO3=2.21−x gm

Moles of CaCO3 decomposed = moles of CaO formed

x100= moles of CaO formed

∴ weight of CaO formed =x100×56
Moles of MgCO3 decomposed = moles of MgO formed

2.21−x84= moles of MgO formed

∴  weight of MgO formed =2.21−x84×40 ⇒2.21−x84×40+x100×56=1.152

∴x=1.187g= weight of CaCO3 and weight of MgCO3=1.023g

Stream:JEESubject:ChemistryTopic:Mole ConceptSubtopic:Miscellaneous/Mixed
⏱ 2mℹ️ Source: PYQ_2024

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