1. Reaction involved

Effervescence is due to the reaction of sodium bicarbonate with acid:

$\mathrm{NaHCO_3 + H^+ \rightarrow CO_2 + H_2O + Na^+}$

Thus, 1 mole of $\mathrm{NaHCO_3}$ produces 1 mole of $\mathrm{CO_2}$.

2. Given volume of $CO_2$

Released $CO_2$ volume:

$0.25\ \text{mL} = 0.25 \times 10^{-3}\ \text{L} = 2.5 \times 10^{-4}\ \text{L}$

3. Calculate moles of $CO_2$

Using molar volume $= 25.0\ \text{L mol}^{-1}$,

$n(\mathrm{CO_2}) = \frac{V}{V_m} = \frac{2.5 \times 10^{-4}}{25.0} = 1.0 \times 10^{-5}\ \text{mol}$

4. Moles of $NaHCO_3$

From stoichiometry,

$n(\mathrm{NaHCO_3}) = n(\mathrm{CO_2}) = 1.0 \times 10^{-5}\ \text{mol}$

5. Mass of $NaHCO_3$ in tablet

$m = nM = (1.0 \times 10^{-5})(84) = 8.4 \times 10^{-4}\ \text{g}$

Convert to mg:

$8.4 \times 10^{-4}\ \text{g} = 0.84\ \text{mg}$

6. Percentage of sodium bicarbonate in 10 mg tablet

$\%\,\mathrm{NaHCO_3} = \frac{0.84}{10} \times 100 = 8.4\%$

7. Option check

• A: $33.6$ ❌
• B: $0.84$ ❌ (this is mass in mg, not percentage)
• C: $8.4$ ✅
• D: $16.8$ ❌

Therefore, the correct answer is Option C.