States of Matter - Result Question 8
8. A $10 mg$ effervescent tablet containing sodium bicarbonate and oxalic acid releases $0.25 mL$ of $CO _2$ at $T=298.15 K$ and $p=1$ bar. If molar volume of $CO _2$ is $25.0 L$ under such condition, what is the percentage of sodium bicarbonate in each tablet?
[Molar mass of $NaHCO _3=84 g mol^{-1}$ ]
(2019 Main, 11 Jan I)
8.4
(b) 0.84
(c) 16.8
(d) 33.6
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Solution:
$2 NaHCO_3+H_2C_2O_4 \longrightarrow 2 CO_2+Na_2C_4O_6+H_2O$
$ 2 mol \quad \quad \quad \quad 1 mol \quad \quad \quad 2 mol $
$\Rightarrow$ In the reaction, number of mole of $CO_2$ produced.
$ \begin{alignedat} n & =\frac{p V}{R T}=\frac{1 \text { bar } \times 0.25 \times 10^{-3} L}{0.0831 \hspace {1mm} L \hspace {1mm} bar \hspace {1mm} K^{-1} mol^{-1} \times 298.15 K} \\ & =1.02 \times 10^{-5} mol \end{aligned} $
Number of mole of $NaHCO _3=\dfrac{\text { Weight of } NaHCO _3}{\text { Molecular mass of } NaHCO _3}$
$\therefore \quad w _{NaHCO _3} =1.02 \times 10^{-5} \times 84 \times 10^{3} mg$
$ = 0.856\ \text{mg}$
$\Rightarrow NaHCO_3$ % $ =\frac{0.856}{10} \times 100=8.56 $%
BETA
