A 10 mg effervescent tablet releases 0.25 mL of CO2 at 298.15 K and 1 bar (molar volume=25.0 L). What is the percentage…

🧠 The "Fizz" stoichiometric link When an effervescent tablet hits water, Sodium Bicarbonate ($\ce{NaHCO3}$) reacts to release $CO_2$ gas in a strict 1:1 ratio. By measuring the volume of the "Fizz" ($0.25\,mL$), we can backtrack to find exactly how much weight of the $10\,mg$ tablet was actually bicarbonate.

🗺️ Strategic Roadmap Step 1: Analyze the Fizz. $V = 0.25\,mL$ at $298\,K$ ($Molar Vol = 25.0\,L$). $\text{Moles } CO_2 = 0.25 \times 10^{-3} / 25.0 = 1 \times 10^{-5}\,mol$

Step 2: Backtrack to Salt. Since ratio is $1:1$: Mass $\ce{NaHCO3} = (1 \times 10^{-5}) \times 84 = 8.4 \times 10^{-4}\,g = 0.84\,mg$.

Step 3: Calculate Purity. Tablet total $= 10\,mg$. $\% = (0.84 / 10) \times 100 = 8.4\%$

⚡ The 5-Second Scan $0.25 / 25 = 0.01$. $0.01 imes 84 = 0.84$. $0.84$ out of $10$ is 8.4%.

$\boxed{\text{Answer: (d) 8.4}}$