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Grade: Upto college level
        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.
10 years ago

Answers : (1)

Pratham Ashish
17 Points
							

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

10 years ago
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