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BITSAT 2024
Exam
BITSAT
Year
2024
Subject
Chemistry
Chapter
Electrochemistry

BITSAT 2024 · Chemistry · Electrochemistry

Calculate the molarity of a solution containing 5 g of NaOH...

Calculate the molarity of a solution containing 5 g of NaOH dissolved in the product of H2−O2\mathrm{H}_2-\mathrm{O}_2 fuel cell operated at 1 A current for 595.1 hours. (Assume F=96500C/mol\mathrm{F}=96500 \mathrm{C} / \mathrm{mol} of electron and molecular weight of NaOH as \left.40 \mathrm{~g} \mathrm{~mol}^{-1}\right).

Options

  1. A0.625 M
  2. B0.05 M
  3. C0.1 M
  4. D6.25 M

Correct answer

A. 0.625 M

Step-by-step solution

The overall reaction of H2−\mathrm{H}_2-O2\mathrm{O}_2 fuel is, 22H2+\mathrm{H}_2+O2\mathrm{O}_2 \longrightarrow 2 H2\mathrm{H}_2O\mathrm{O}ThusThus4 F=\mathrm{ F}=4 \times 96500 C\mathrm{C} produce water =2=2mol=\mathrm{ mol}=3636g\mathrm{ g} Charge actually passed =i×t=1×595.1×60×60=2142360C\begin{aligned} \text{ Charge actually passed } & =i \times t \\ & =1 \times 595.1 \times 60 \times 60 \\ & =2142360 \mathrm{C} \end{aligned}∴ Water produced =364×96500×2142360≅200g or 200mL∴ Molarity of NaOH solution =540×1200×1000=0.625M\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}
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