An impure sample of weighing 0.588 g was dissolved in water. Excess and were added to it. The libera — Redox Reactions and Volumetric Analysis Chemistry Question
Question
An impure sample of $K_2Cr_2O_7$ weighing 0.588 g was dissolved in water. Excess $KI$ and $HCl$ were added to it. The liberated $I_2$ required exactly 12 mL of 0.5 N $Na_2S_2O_3$ solution to reach the end point. What is the percentage purity of the $K_2Cr_2O_7$ sample? (Molar mass of $K_2Cr_2O_7$ = 294 g/mol)
Answer: B
💡 Solution & Explanation
Using Law of Equivalence: Equivalents of $K_2Cr_2O_7$ = Equivalents of $Na_2S_2O_3$. Eq = $N \times V(in L) = 0.5 \times (12/1000) = 0.006$ eq. Equivalent weight of $K_2Cr_2O_7 = MW / 6 = 294 / 6 = 49$. Pure mass = $0.006 \times 49 = 0.294$ g. Percentage purity = $(0.294 / 0.588) \times 100 = 50\%$.
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