Calculate volume strength of H 2 O 2 used. | TestHub

Passage / Comprehension

$H_{2} O_{2} + 2 KI 40% yield I_{2} + 2 KOH H_{2} O_{2} + 2 KMnO_{4} + 3 H_{2} SO_{4} 50% yield K_{2} SO_{4} + 2 MnSO_{4} + 3 O_{2} + 4 H_{2} O$

100 ml of $H_{2} O_{2}$ sample was divided into two parts. First part was treated with KI and KOH formed required 200 ml of $M /2 H_{2} SO_{4}$ for complete neutralisation. Other part was treated with just sufficient $KMnO_{4}$ yielding 6.72 L of $O_{2}$ at 1 atm & 273 K.

Calculate volume strength of $H_{2} O_{2}$ used.

Options:

Answer:

Solution:

Moles of KOH neutralized $= 2 \times 0.2 L \times 0.5 M = 0.2 m o l$ ; thus, moles of H₂O₂ ( $n_{1}$ ) $= \frac{0.2}{2 \times 0.40} = 0.25 m o l$ .

Moles of O₂ produced $= \frac{6.72 L}{22.4 L / m o l} = 0.3 m o l$ ; using the given stoichiometry and $50%$ yield, moles of H₂O₂ ( $n_{2}$ ) $= \frac{0.3}{3 \times 0.50} = 0.20 m o l$ .

Total Concentration: Total moles $= 0.25 + 0.20 = 0.45 m o l$ in $100 m L$ ( $0.1 L$ ), giving a Molarity of $\frac{0.45}{0.1} = 4.5 M$ .

Volume Strength: Volume strength $=$ Molarity $\times 11.35 = 4.5 \times 11.35 = 51.08$ (Option A).

Chemistry - Concentration Terms Question with Solution | TestHub

Options:

Doubts & Discussion