In water, 0 . 5 g of an impure sample of oxalate was dissolved, making the solution to 100 ml . 10 ml of this required 15 ml of N / 20 KMnO 4 solution on titration. Calculate the percentage of pure oxalate in the sample.

A solution of 15 ml of N / 20 KMnO 4 will have 15 / 20 milliequivalents of KMnO 4 No. of milliequivalents of Oxalate in 10 ml solution = 3 / 4 No. of milliequivalents of Oxalate in 100 ml solution = 30 / 4 Valence Factor of Oxalate = 2 Number of milliequivalents = number of millimoles × valence factor Number of millimoles of oxalate = 30 8 millimoles The molar mass of oxalate, C 2 O 2 2 - = 88 g mol - 1 Weight in grams of oxalate = 30 × 88 8 × 1000 g = 0 . 33 g % Pure oxalate = 0 . 33 0 . 5 × 100 = 66 %

Both Cr 2 O 7 2 - ( aq ) and MnO 4 - ( aq ) can be used to titrate Fe 2 + ( aq ) . If in a given titration, 24 . 50 cm 3 of 0 . 1 M Cr 2 O 7 2 - were used, then what volume of 0 . 1 M MnO 4 - solution will be consumed for the same titration in acidic medium?

Molarity of Cr 2 O 7 2 - = 0 . 1 M Molarity of MnO 4 - = 0 . 1 M Volume of Cr 2 O 7 2 - = 24 . 50 ml Volume of MnO 4 - needs to be calculated, Applying law of equivalence for titration between Fe 2 + and Cr 2 O 7 2 - Milliequivalents of Cr 2 O 7 2 - = Milliequivalents of Fe 2 + Milliequivalents = millimoles × valency factor Millimoles = Molarity × Volume in ml Millimoles of Cr 2 O 7 2 - = 0 . 1 × 24 . 5 Valency factor of Cr 2 O 7 2 - = 6 Milliequivalents of Cr 2 O 7 2 - = 0 . 1 × 24 . 5 × 6 = 14 . 7 Milliequivalents of Fe 2 + = 14 . 7 These much Fe 2 + will react with M n O 4 - Applying law of equivalence for titration between Fe 2 + and MnO 4 - Milliequivalents of MnO 4 - = Milliequivalents of Fe 2 + Milliequivalents of MnO 4 - = 14 . 7 Millimoles of MnO 4 - × Valency factor of MnO 4 - = 14 . 7 Valency Factor of MnO 4 - = 5 in acidic medium Millimoles of MnO 4 - = 14 . 7 5 = 2 . 94 Millimoles of MnO 4 - =Molarity of MnO 4 - × Volume of MnO 4 - in ml 2 . 94 = 0 . 1 × Volume of MnO 4 - Volume of MnO 4 - = 29 . 4 ml