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Equivalent Mass

4.08 gm of mi...

Question

4.08 gm of mixture BaO (Ba - 137) and unknown carbonate ${MCO}_{3}$ was heated strongly. the residue weighed 3.96 gm. the residue was dissolved in 100 ml of 1N HCl. The excess required 16 ml of 2.5 N NaOH for complete neutralization. Find which element is M.

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Solution

Given $B a O + M C O_{3} = 4.08 g m$
Now, when $M C O_{3} (1 m o l e) Δ - \to M O 1 m o l e + C O_{2} 1 m o l e$
As metal carbonate will loose its weight correspondingly to the weight of $C O_{2}$ . So, weigh loss is given by $\Rightarrow 4.08 - 3.96 = 0.12 g m$
$\Rightarrow 0.12 g m$ of $C O_{2}$ is produced.
$1$ mole of $C O_{2} = 44 g$
$\Rightarrow 1 g = \frac{1}{44}$ mole
$\Rightarrow 0.12 g = \frac{0.12}{44}$ mole= $0.0027$ mole of $C O_{2}$ produced
Now, $1$ mole of $M C O_{3}$ produce $1$ mole of $C O_{2}$ . Vice versa, $0.0027$ mole of $C O_{2}$ is produced by $0.0027$ mole of $M C O_{3}$ also, which will yield $0.0027$ mole of $M O$ .

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