A 10 mg effervescent tablet contianing sodium bicarbonate and oxalic acid releases 0.25 ml of CO₂ at T = 298.15 K and p = 1 bar. If molar volume of CO₂ is 25.0L under such condition, what is the percentage of sodium bicarbonate in each tablet ? [Molar mass of NaHCO₃ = 84 g mol^-1]

$\begin{array}{l}2{\mathrm{NaHCO}}_3+{\mathrm{H}}_2{\mathrm{C}}_2{\mathrm{O}}_4\to 2{\mathrm{CO}}_2+{\mathrm{Na}}_2{\mathrm{C}}_2{\mathrm{O}}_4+2{\mathrm{H}}_2\mathrm{O}\end{array}$

Here, number of moles of CO₂ = $\frac{0.25\times {10}^{-3}}{25}={10}^{-5}$

Now, one moles of CO₂ is produced by one mole of NaHCO₃.

$\therefore$the number of moles of NaHCO₃ in the given reaction= number of moles of CO₂ = 10^-5

Now, the weight of NaHCO₃ = 10^-5 x 84 = 84 x 10^-5g

$\therefore \% \mathrm{Mass}=\frac{84\times {10}^{-5}}{10\times {10}^{-3}}\times 100=8.4\%$