Question

# 1.0 g of a sample containing KMnO4 and some inert impurity was dissolved in dilute solution of KOH and volume was made up to 100 mL. To this solution, some ammonia gas was passed where the following redox reaction occured: NH3+KMnO4→ N2+MnO2(s) Volume of N2 measured at N.T.P. was 28 mL. The solution was filtered off to remove MnO2(s) and a 10 mL portion of the filtrate required 22.5 mL of a 0.01 M oxalate solution to reach the end point. The mass percentage of KMnO4 in the original sample is .

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Solution

## Number millimoles of N2 produced =2822.4=1.25 mmol The balanced chemical reaction between ammonia and KMnO4 is: 2NH3+2KMnO4→N2+2MnO2+2KOH+2H2O The amount of ammonia passed =2.5 mmol ⇒ammonia is the limiting reagent as the solution is further titrated against oxalate solution to reduce the left over permanganate. Number of milli moles of KMnO4 reacted with NH3 = 2.5 Total number of milli equivalents of oxalate consumed=22.5×0.01×10×2=4.5 Milli equivalents of KMnO4 left unreacted after reaction with ammonia = 4.5 Number of milli moles of KMnO4=4.53=1.5 Total number of milli moles of KMnO4=2.5+1.5=4 ⇒Mass of KMnO4=4×158×10−3=0.632 Mass % = 0.6321×100=63.2%

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