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Question: Answered & Verified by Expert
20.0 g of a magnesium carbonate sample decomposes on heating to give carbon dioxide and 8.0 g magnesium oxide. What will be the percentage purity of magnesium carbonate in the sample?

(Atomic weight: Mg = 24)
ChemistrySome Basic Concepts of ChemistryNEETNEET 2015 (Phase 2)
Options:
  • A 75
  • B 96
  • C 60
  • D 84
Solution:
1908 Upvotes Verified Answer
The correct answer is: 84
MgCO3sMgOs+CO2g

Moles of MgCO3=2084=0.238 mol

From above equation.

1 mole MgCO3 gives 1 mole MgO

0.238 mole MgCO3 will give 0.238 mole MgO

=0.238×40 g=9.523 g MgO

Practical yield of MgO=8 g MgO

% Purity=89.523×100=84%

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