g of an oxalate was dissolved in water and the solution made to mL. mL of this solution required mL — Redox Reactions and Volumetric Analysis Chemistry Question
Question
$0.5$ g of an oxalate was dissolved in water and the solution made to $100$ mL. $10$ mL of this solution required $8$ mL of $N/20$ $KMnO_4$ for oxidation. The percentage of oxalate ($C_2O_4^{2-}$) in the sample is:
Answer: A
💡 Solution & Explanation
Meq in $10$ mL = $8 \times 0.05 = 0.4$. <br>Meq in $100$ mL = $4$. <br>Mass of oxalate = $(\text{Meq} \times E) / 1000 = (4 \times 44) / 1000 = 0.176$ g. <br>$\% = (0.176 / 0.5) \times 100 = 35.2\%$.
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