Chemistry - Concentration Terms Question with Solution | TestHub

ChemistryConcentration TermsVolume StrengthMedium2 minQB

ChemistryMediumsingle choice

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 moles of KOH produced :

Options:

Answer:

Solution:

Neutralization Reaction: $2 K O H + H_{2} S O_{4} \to K_{2} S O_{4} + 2 H_{2} O$

Moles of H₂SO₄: $M o l a r i t y \times V o l u m e (L) = 0.5 M \times 0.2 L = 0.1 m o l e$

Mole Relationship: From the balanced equation, $1 m o l e$ of H₂SO₄ neutralizes $2 m o l es$ of KOH.

Calculation: Moles of KOH produced $= 2 \times 0.1 = 0.2$ (Hence, Option B is correct).

Stream:JEESubject:ChemistryTopic:Concentration TermsSubtopic:Volume Strength

⏱ 2mℹ️ Source: QB

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