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One gram of carbonate $(M_{2} C O_{3})$ on treatment with excess HCl produces 0.01186 mole of $C O_{2} .$ The molar mass of $M_{2} C O_{3}$ in gm/mole is

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84.3

118.6

11.86

1186

answer is C.

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Detailed Solution

$M_{2} C O_{3} + 2 H C l \to 2 M C l + C O_{2} + H_{2} O$

$1 gm M_{2} C O_{3}____0.01186 m o l e s C O_{2}$

?____________ $1 mole {CO}_{2}$

$\frac{1 x1}{0.01186} = 84.3$

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