 # 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. 12 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

2 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 ml
Weight of pure kmno4 :- 90/100 ×3.53 g = 3.177 g
Moles of kmno4 :- 3.177/158
Conc. 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=3.177/158×5×20/1000
We get X equal to 5.88 which is nearly equal to "6"
(I am just a student in 12th !!!)