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Laws of Mass Conservation

1 gram of a m...

Question

1 gram of a metal carbonate $(M_{2} C O_{3})$ on treatment with excess $H C l$ produces 0.01186 mole of $C O_{2}$ . The molar mass of $(M_{2} C O_{3})$ in $g m o l^{- 1}$ is

1186

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84.3

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118.6

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11.86

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Solution

The correct option is B $84.3$
$M_{2} C O_{3} + 2 H C l \to 2 M C l + H_{2} O + C O_{2}$
$0.01186 moles C O_{2} = 0.01186 moles of M_{2} C O_{3} = 1 g M_{2} C O_{3}$
$Molar mass of M_{2} C O_{3} = \frac{Mass of M_{2} C O_{3}}{No. of moles of M_{2} C O_{3}} = \frac{1 g}{0.01186 m o l} = 84.3 g / m o l$

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