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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
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