1.6g of pyrolusite ore was treated with 50cc of 1 N oxalic acid and some sulphuric acid. The oxalic aci left unecomposed was raised to 250cc in a flask. 25 cc of this solution when treated with 0.1 N KMnO4 sol. required 32 cc of the sol. Find out the % purity of the ore and also the percentage of oxygen available.
Equivalents of Oxalix acid = NV = 50 milli eq
Now lets go the end of the question
Equivalents of KMnO4= NV = 3.2 milli eq = Equivalents of Oxalic acid
So equivalents of Oxalic Acid left undecomposed after step 1 = 32 milli eq
Equivalents of Oxalic Acid reacted = 18 milli eq
eq of MnO2 reacted = 18 milli eq
mol of MnO2 reacted = 9 milli mole
wt of MnO2 = 9 x 87/1000 grm = 0.783 gm
% purity = ~48.9 %
Mona's soln is perfect...
Not sure what percent of oxygen available means but I think it means the wt of oxygen in the ore/ wt of ore..
That would be 0.28/1.6 x100 = 17.5 %
Moderation Team
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