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```
9.8g of FeSO4.(NH4)2SO4.xH20 was dissolved in 250ml of its solution. 20ml of this solution required 20ml of Pottassiumpermanganate (KMnO4) SOLUTION CONTAIONING 3.53g of 90% pure KMnO4 dissolved per litre. Calculate x ? Ans (6) Sir, it is the question of "Modern Approach To chemical calculations" by Ramendra c. Mukerjee, page no.154, Q.no.35.

```
11 years ago

```							conc of KMnO4 solu. = 3.53* 0.9  gm / lt.
=3.177 gm /lt
no. of equivalents of KMnO4  in 20 ml = 3.177  * 0.02 / 31.6
( eq. wt . of  KMnO4  is 31.6 )
= 2.01 m eq.
it reacts with the 20 ml . of given solu. , so the given solu. must also have the same no. of milli equivalents
no. of milli equivalents of FeSO4 . (NH4) 2 SO4 . x H2O  in this 20 ml = 2.01
total no. of milli equivalents  in 250 ml . = 250 * 2.01 /20
= 25.125
for FeSO4 . (NH4) 2 SO4 . x H2O  change in oxidation state in aredox reaction is +1
so,  the molecular wt . & equivalent wt. will be equal
so,  total no. of milli moles  in 250 ml  = 25.125
25.125 milli moles are presnt in the 9.8 gm , so
the mol. wt . of FeSO4 . (NH4) 2 SO4 . x H2O   =   9.8 / 25.125 *10-3
= 390.05 gm.
from the mol formula ,
mol. wt = 390.05 =  284 + x 18
x = 106.05 /18
= 5.88
this shows that the x must be 6
```
11 years ago
```							Moles of the salt :- 9.8/(284+18x)( Let the no. of h2o be X)N-factor of salt is 1 as only Fe changes its oxidation state from +2 to +3 Molarity of salt :-( 9.8/(284+18x))×4(as it is dissolved in 250 ml solution)Volume of salt taken is 20 mlWeight of pure kmno4 :- 90/100 ×3.53 g = 3.177 gMoles of kmno4 :- 3.177/158Conc. of kmno4 :- (3.177/158)/1(As weight of kmno4 is given in per litre)N-factor of kmno4 :- 5 Vol. Of kmno4 taken :- 20 ml By equating no. of equivalents (Formula used :- N1V1=N2V2=no.of equivalents and N=M×N-factor and where N is normality,M is mooarity and V is volume)(9.8/284+X)×4×20/1000×1=3.177/158×5×20/1000We get X equal to 5.88 which is nearly equal to "6"(I am just a student in 12th !!!)
```
6 months ago
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