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NEET CHEMISTRYSome Basic Concepts of ChemistryMedium

Question

20.0 g20.0\text{ g} of a magnesium carbonate sample decomposes on heating to give carbon dioxide and 8.0 g8.0\text{ g} magnesium oxide. What will be the percentage purity of magnesium carbonate in the sample? (Atomic weight of Mg = 24)

A

75

B

96

C

60

D

84

Step-by-Step Solution

The balanced chemical equation for the decomposition of magnesium carbonate is: MgCO3(s)ΔMgO(s)+CO2(g)\text{MgCO}_3(s) \xrightarrow{\Delta} \text{MgO}(s) + \text{CO}_2(g) From the stoichiometry of the reaction, 1 mole1\text{ mole} of MgCO3\text{MgCO}_3 yields 1 mole1\text{ mole} of MgO\text{MgO}. Molar mass of MgCO3=24+12+(3×16)=84 g/mol\text{MgCO}_3 = 24 + 12 + (3 \times 16) = 84\text{ g/mol}. Molar mass of MgO=24+16=40 g/mol\text{MgO} = 24 + 16 = 40\text{ g/mol}. According to the equation, 40 g40\text{ g} of MgO\text{MgO} is obtained from 84 g84\text{ g} of pure MgCO3\text{MgCO}_3. Therefore, 8.0 g8.0\text{ g} of MgO\text{MgO} will be obtained from =(8440)×8.0=16.8 g= \left(\frac{84}{40}\right) \times 8.0 = 16.8\text{ g} of pure MgCO3\text{MgCO}_3. The mass of pure MgCO3\text{MgCO}_3 in the sample is 16.8 g16.8\text{ g}. The total mass of the impure sample is given as 20.0 g20.0\text{ g}. Percentage purity =(Mass of pure MgCO3Total mass of the sample)×100=(16.820.0)×100=84%= \left(\frac{\text{Mass of pure MgCO}_3}{\text{Total mass of the sample}}\right) \times 100 = \left(\frac{16.8}{20.0}\right) \times 100 = 84\%.

Exam Context & Concepts Covered

This question aligns with the NEET CHEMISTRY syllabus, specifically targeting concepts from Some Basic Concepts of Chemistry. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

CHEMISTRYSome Basic Concepts of Chemistrymagnesiumcarbonatesampledecomposesheating

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