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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 ?

(At. Wt. : Mg = 24)

  • Option 1)

    75

  • Option 2)

    96

  • Option 3)

    60

  • Option 4)

    84

 

Answers (1)

As learnt in concept

Number of Moles -

No of moles = given mass of substance/ molar mass of substance

-

 

 MgCO_{3} \rightarrow MgO(s)+CO2(g)

supposed meals of = \frac{mass \:\:in\:\: grams}{molar \:\:mass}

 

 = \frac{20}{84} = 0.238 \:\:mal

By the stoichoimetry of the reaction alone, we know that 1 mal of the MgCO_{3} gives 1 mal of MgO moles of MgO obtained 

= \frac{8}{40} = 0.2 moles of MgO that should have been obtained in a pure sample = 0.238

\therefore moles of MgCO_{3} in the sample

= moles of MgO in the sample

= 0.2

\therefore % purity = \frac{0.2}{0.238} = 84\%

 


Option 1)

75

This solution is incorrect.

Option 2)

96

This solution is incorrect.

Option 3)

60

This solution is incorrect.

Option 4)

84

This solution is correct.

Posted by

Vakul

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