0.3 g of an oxalate salt was dissolved in 100 mL of water. The solution required 90 mL of N/20 for c — Redox Reactions and Volumetric Analysis Chemistry Question
Question
0.3 g of an oxalate salt was dissolved in 100 mL of water. The solution required 90 mL of N/20 $KMnO_4$ for complete oxidation. The mass percentage of oxalate ion in the salt is:
Answer: B
💡 Solution & Explanation
Milli-equivalents of $KMnO_4$ used = $90 \times (1/20) = 4.5$ meq = $0.0045$ equivalents. By equivalence, eq of oxalate = $0.0045$ . Equivalent weight of $C_2O_4^{2-}$ = Molar mass / 2 = 88 / 2 = 44 g/eq. Mass of oxalate = $0.0045 \times 44 = 0.198$ g. Percentage = $(0.198 / 0.3) \times 100 = 66\%$ .
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