Q48 / 180 · NEET 2015 (Phase 2) · Chemistry

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)

  • A
    75
  • B
    96
  • C
    60
  • Correct answer
    D
    84

Solution

MgCO3s⟶MgOs+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%

Hence, the correct answer is option: 'D'.

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