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Question

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+I2Na2S4O6+2NaI
Find the value of X. (The molecular weight of  KMnO4=158 gmol1)  
. 


Solution

According to question, meq. of KMnO4 = meq. of I2= meq. of hypo = 50×0.1×1=5 
X1585×103 = 5
X=0.158 g 
Thus, mass of KMnO4 is 0.158 g

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