Question

A $10\, mg$ effervescent tablet contianing sodium bicarbonate and oxalic acid releases $0.25\, ml$ of ${CO_2}$ at $T$ = $298.15\, K$ and $p = 1$ bar. If molar volume of ${CO_2}$ is $25.0\, L$ under such condition, what is the percentage of sodium bicarbonate in each tablet ? [Molar mass of ${NaHCO_3 = 84 \; g \; mol^{-1}}$]

• A. 16.8
• B. 8.4
• C. 0.84
• D. 33.6

Answer 1

The Correct Option is B

$2 NaHCO _{3}+ H _{2} C _{2} O _{4} \rightarrow Na _{2} C _{2} O _{4}+2 CO _{2}+2 H _{2} O$
Here, number of moles of $CO _{2}=\frac{0.25 \times 10^{-3}}{25.9} \approx 10^{-5}$
Now, one mole of $CO _{2}$ is produced by one mole of $NaHCO _{3}$.
$\therefore$ the number of moles of $NaHCO _{3}$ in the given reaction $=$ number of moles of $CO _{2}= 10^{-5}$
Now, the weight of $NaHCO _{3}=10^{-5} \times 84$
$=84 \times 10^{-5} g$
$\therefore \%$ Mass $=\frac{84 \times 10^{-5}}{10 \times 10^{-3}} \times 100$
$=8.4 \%$