g of an impure sample of on reaction with excess liberated which was passed into solution. The liber — Redox Reactions and Volumetric Analysis Chemistry Question
Question
$0.5$ g of an impure sample of $MnO_2$ on reaction with excess $HCl$ liberated $Cl_2$ which was passed into $KI$ solution. The $I_2$ liberated required $40$ mL of $0.1 N$ $Na_2S_2O_3$. The percentage purity of $MnO_2$ is ($Mw = 87$):
Answer: A
💡 Solution & Explanation
Meq $MnO_2$ = Meq $Na_2S_2O_3 = 40 \times 0.1 = 4$. <br>Mass of $MnO_2 = (4 \times (87/2)) / 1000 = 0.174$ g. <br>$\% = (0.174 / 0.5) \times 100 = 34.8\%$.
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