X g of KMnO4 on being treated with KI in acid solution liberates just sufficient I2 to react with 50 ml of 0.1 M hypo solution. Na2S2O3+I2→Na2S4O6+2NaI Find the value of X. (The molecular weight of KMnO4=158gmol−1) .
Open in App
Solution
According to question, meq. of KMnO4 = meq. of I2= meq. of hypo = 50×0.1×1=5 X1585×10−3 = 5 X=0.158g Thus, mass of KMnO4 is 0.158 g