Standardization of KMnO4 Trial 1 Trial 2 1. Mass of flask +...

Standardization of KMnO4 Trial 1 Trial 2 1. Mass of flask + sodium oxalate, g: 123.531 2. Mass of empty flask, g: 123.184 3. Mass of sodium oxalate, g: 0.347 4. Initial buret reading of KMnO4, ml: 1.8 5. Final buret reading of KMnO4, ml: 30.2 6. Volume of KMnO4 used, in ml: 28.4 7. Volume of KMnO4 used, in L: .0284 L 8. Molarity of MnO4- solution, M: .035 M 9. Average molarity of MnO4- solution: .035 M Determination of Concentration of Unknown Oxalate Solution 10. Volume of unknown oxalate solution, l: 25 11. Initial buret reading of KMnO4, ml: 0.2 12. Final buret reading of KMnO4, ml: 17.1 13. Volume of KMnO4 used, in ml: 16.9 14. Volume of KMnO4 used, in L: .0169 L 15. Molarity of unknown oxalate solution, M: 16. Average concentration of oxalate solution, M:

Transcript

00:01 They have given the certain data that is standardization of kmno4 so this is the data they have given standardization of kmn values they have given that is mass of flask plus sodium oxalate that is equal to 123 .3 .31 grams so mass of empty flask that is in grams 123 .184 mass of sodium oxalate ingram 0 .347 grams so initial bureate reading of km and o4 that is 1 .8 ml final bureate reading of k mn .o4 30 .2m.

00:33 Volume of kmno4 used 28 .4m.

00:36 Volume of kmno4 used in liters, 0 .0284 liters.

00:43 We have to find out the molarity of mno4 minus solution, molars and average molarity of kmno4 minus solution.

00:51 So let us see here.

00:53 So that is equivalence.

00:56 So equivalence of sodium oxalate.

01:00 It must be equal it is equal to so equivalence of sodium oxalate is equal to equivalence of kmno4 equivalence of kmno4 so that means here so equivalence of kmno4 that means here mole of sodium oxalate moles of sodium oxalate so here first of all we have to find out equivalence of sodium oxalate how to find out mole of sodium oxalate, oxalate into valency factor of sodium oxalate.

01:41 So what will be this one? so what is the moles 0 .347? okay, number of moles is calculated by weight by molar mass.

01:53 So molar mass it is 134 into what is the valency factor, valency factor it is 2.

02:00 So this gives the number of moles weight by molar mass.

02:06 So whenever we calculate this, it is equivalent to 0 .005, that is equation 1.

02:13 So then coming to the equivalence of kmno4, we have to find out.

02:20 So how to find out equivalence of km and o4? the equivalence of kmno4 is equal to molarity of kmno4.

02:29 Molarity of km and o4 and into volume used into volume used into into valency factor because here the number of moles that is equal to molarity into volume so molarity into volume so what is the molarity molarity it is we don't know into valency factor into volume used into valency factor volume used zero 0 .0 .0 .0 .084.

03:12 0284 into 5.

03:14 So because it should be taken in the liters, whenever we calculate this, we'll get the equivalence factor of km and 4.

03:22 What is that? that is equal to? so here m, it is one variable.

03:29 So we can't do it.

03:31 So let it be equation two.

03:33 Okay...

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