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pikachu singh 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.
8 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

8 years ago
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