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JEE Main levelSingle correct optionDifficulty 2 of 3mt078

Pure dolomite, $CaCO X_{3} \cdot MgCO X_{3}$ (Ca 40, Mg 24, C 12, O 16), is calcined completely to $CaO$ and $MgO$ . The percentage loss in mass of the solid is closest to

A $23.9%$
B $52.2%$
C $47.8%$
D $44.0%$

Answer: C

CaMg (CO X_{3}) X_{2} CaO + MgO + 2 CO X_{2}

. Molar mass of dolomite

= 184 g/mol

and two moles of

CO X_{2}

weigh

88 g

. Loss

= 88/184 \times 100 = 47.8%

.

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