# 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.

Pratham Ashish
17 Points
14 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

Satyam
13 Points
3 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 !!!)