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A $10 \mathrm{mg}$ effervescent tablet containing sodium bicarbonate and oxalic acid releases $0.25 \mathrm{~mL}$ of $\mathrm{CO}_{2}$ at $\mathrm{T}=298.15 \mathrm{~K}$ and $\mathrm{P}=1$ 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 $\left.\mathrm{NaHCO}_{3}=84 \mathrm{~g} \mathrm{~mol}^{-1}\right]$
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2652 Upvotes
Verified Answer
The correct answer is:
8.4
$$
2 \mathrm{NaHCO}_{3}+\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4} \longrightarrow \mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}+2 \mathrm{CO}_{2}+2 \mathrm{H}_{2} \mathrm{O}
$$
Moles of $\mathrm{CO}_{2}$ evolved $=\frac{0.25}{25 \times 10^{3}}=10^{-5}$
$$
\begin{aligned}
\therefore \quad & \text { Moles of } \mathrm{NaHCO}_{3}=10^{-5} \\
\therefore \quad \text { Mass of } \mathrm{NaHCO}_{3} &=84 \times 10^{-5} \mathrm{~g} \\
&=0.84 \times 10^{-3} \mathrm{~g}=0.84 \mathrm{mg}
\end{aligned}
$$
$\therefore \quad \%$ by weight $=\times 100=8.4 \%$
2 \mathrm{NaHCO}_{3}+\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4} \longrightarrow \mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}+2 \mathrm{CO}_{2}+2 \mathrm{H}_{2} \mathrm{O}
$$
Moles of $\mathrm{CO}_{2}$ evolved $=\frac{0.25}{25 \times 10^{3}}=10^{-5}$
$$
\begin{aligned}
\therefore \quad & \text { Moles of } \mathrm{NaHCO}_{3}=10^{-5} \\
\therefore \quad \text { Mass of } \mathrm{NaHCO}_{3} &=84 \times 10^{-5} \mathrm{~g} \\
&=0.84 \times 10^{-3} \mathrm{~g}=0.84 \mathrm{mg}
\end{aligned}
$$
$\therefore \quad \%$ by weight $=\times 100=8.4 \%$
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