Q17 / 100 · BITSAT 2024 (Memory Based Paper 3) · Chemistry
Calculate the molarity of a solution containing 5 g of NaOH dissolved in the product of $\mathrm{H}_2-\mathrm{O}_2$ fuel cell operated at 1 A current for 595.1 hours. (Assume $\mathrm{F}=96500 \mathrm{C} / \mathrm{mol}$ of electron and molecular weight of NaOH as $\left.40 \mathrm{~g} \mathrm{~mol}^{-1}\right)$.
- Correct answerA0.625 M
- B0.05 M
- C0.1 M
- D6.25 M
Solution
The overall reaction of $\mathrm{H}_2-\mathrm{O}_2$ fuel is,
$2 \mathrm{H}_2+\mathrm{O}_2 \longrightarrow 2 \mathrm{H}_2 \mathrm{O}$
Thus $4 \mathrm{~F}=4 \times 96500 \mathrm{C}$ produce water $=2 \mathrm{~mol}=36 \mathrm{~g}$
$\begin{aligned}
\text { Charge actually passed } & =i \times t \\
& =1 \times 595.1 \times 60 \times 60 \\
& =2142360 \mathrm{C}
\end{aligned}$
$\begin{aligned}
& \therefore \text { Water produced }=\frac{36}{4 \times 96500} \times 2142360 \\
& \cong 200 \mathrm{~g} \text { or } 200 \mathrm{~mL} \\
& \therefore \text { Molarity of } \mathrm{NaOH} \text { solution }=\frac{5}{40} \times \frac{1}{200} \times 1000 \\
& =0.625 \mathrm{M}
\end{aligned}$
Hence, the correct answer is option: 'A'.