States of Matter 1 Question 8
8. A $10 \mathrm{mg}$ effervescent tablet containing sodium bicarbonate and oxalic acid releases $0.25 \mathrm{~mL}$ of $\mathrm{CO}{2}$ at $T=298.15 \mathrm{~K}$ and $p=1$ bar. If molar volume of $\mathrm{CO}{2}$ is $25.0 \mathrm{~L}$ under such condition, what is the percentage of sodium bicarbonate in each tablet?
[Molar mass of $\mathrm{NaHCO}_{3}=84 \mathrm{~g} \mathrm{~mol}^{-1}$ ]
(2019 Main, 11 Jan I)
(a) 8.4
(b) 0.84
(c) 16.8
(d) 33.6
Show Answer
Solution:
- $2 \mathrm{NaHCO}{3}+\mathrm{H}{2} \mathrm{C}{2} \mathrm{O}{4} \longrightarrow 2 \mathrm{CO}{2}+\mathrm{Na}{2} \mathrm{C}{4} \mathrm{O}{4}+\mathrm{H}_{2} \mathrm{O}$
$$ 2 \mathrm{~mol} \quad 1 \mathrm{~mol} \quad 2 \mathrm{~mol} $$
$\Rightarrow$ In the reaction, number of mole of $\mathrm{CO}_{2}$ produced.
$$ \begin{aligned} n & =\frac{p V}{R T}=\frac{1 \text { bar } \times 0.25 \times 10^{-3} \mathrm{~L}}{0.082 \mathrm{~L} \mathrm{~atm} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} \times 298.15 \mathrm{~K}} \ & =1.02 \times 10^{-5} \mathrm{~mol} \end{aligned} $$
Number of mole of $\mathrm{NaHCO}{3}=\frac{\text { Weight of } \mathrm{NaHCO}{3}}{\text { Molecular mass of } \mathrm{NaHCO}_{3}}$
$$ \begin{aligned} \therefore \quad w_{\mathrm{NaHCO}{3}} & =1.02 \times 10^{-5} \times 84 \times 10^{3} \mathrm{mg} \ & =0.856 \mathrm{mg} \ \Rightarrow \mathrm{NaHCO}{3} % & =\frac{0.856}{10} \times 100=8.56 % \end{aligned} $$
