An impure sample of potassium metavanadate, KVO3\text{KVO}_3KVO3, weighing 0.160 g0.160\text{ g}0.160 g is dissolved in dilute sulfuric acid to form a solution containing aqueous VO2+\text{VO}_2^+VO2+ ions. All of the vanadium is converted to VO2+\text{VO}_2^+VO2+ ions. These VO2+\text{VO}_2^+VO2+ ions are reduced to aqueous V2+\text{V}^{2+}V2+ ions by reaction with an excess of zinc according to the equation:
2VO2+(aq)+8H+(aq)+3Zn(s)→3Zn2+(aq)+2V2+(aq)+4H2O(l)2\text{VO}_2^+(\text{aq}) + 8\text{H}^+(\text{aq}) + 3\text{Zn}(\text{s}) \rightarrow 3\text{Zn}^{2+}(\text{aq}) + 2\text{V}^{2+}(\text{aq}) + 4\text{H}_2\text{O}(\text{l})2VO2+(aq)+8H+(aq)+3Zn(s)→3Zn2+(aq)+2V2+(aq)+4H2O(l)
The excess zinc is removed by filtration and washed. The filtrate, which contains the V2+\text{V}^{2+}V2+ ions, is titrated with a 0.0150 mol dm−30.0150\text{ mol dm}^{-3}0.0150 mol dm−3 solution of acidified KMnO4\text{KMnO}_4KMnO4. Exactly 39.40 cm339.40\text{ cm}^339.40 cm3 of the KMnO4\text{KMnO}_4KMnO4 solution are required to oxidise all the V2+\text{V}^{2+}V2+ ions back to VO2+\text{VO}_2^+VO2+ ions. The ionic equation for the reaction of MnO4−\text{MnO}_4^-MnO4− ions with V2+\text{V}^{2+}V2+ ions is:
3MnO4−(aq)+5V2+(aq)+4H+(aq)→3Mn2+(aq)+5VO2+(aq)+2H2O(l)3\text{MnO}_4^-(\text{aq}) + 5\text{V}^{2+}(\text{aq}) + 4\text{H}^+(\text{aq}) \rightarrow 3\text{Mn}^{2+}(\text{aq}) + 5\text{VO}_2^+(\text{aq}) + 2\text{H}_2\text{O}(\text{l})3MnO4−(aq)+5V2+(aq)+4H+(aq)→3Mn2+(aq)+5VO2+(aq)+2H2O(l)
Calculate the percentage purity of the KVO3\text{KVO}_3KVO3 sample. Give your answer to 3 significant figures.
(Use the following relative atomic masses: K=39.1\text{K} = 39.1K=39.1, V=50.9\text{V} = 50.9V=50.9, O=16.0\text{O} = 16.0O=16.0)