Please help with the question attached below:

 
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Molar mass of Mohrs salt (NH4)2Fe(SO4)2 ? 6 H2O is 392,13 g/mol
1 ml of 0,1 N = 0,02 M KMnO4  is equal O,1 M Mohrs salt
50 ml is then equal 1,96 g.
2,5 g were weight before, the ratio is then 1,96 g/2,5 g = 0,784 equal 78,4%

Quote from: aarna on January 26, 2023, 12:58:19 PM
Please help with the question attached below:

Formula of Mohr's salt is FeSO₄.(NH₄)₂SO₄.6H₂O
During redox titrations, only Fe²⁺ oxidized to Fe³⁺
5 electrons  + MnO₄^- ----> Mn ²⁺
5 Fe²⁺------> 5Fe³⁺   + 5 electrons
Equivalents of MnO4 - = N * V(L) = 0.0050 = equivalents of Fe²⁺
Equivalent of Fe²⁺ are moles of Fe²⁺ .
Moles of Fe²⁺ are moles of Mohr's salt = 0.0050 moles
mass of Mohr's salt in 2.5 g of sample = 0.0050 * 392 = 1.96 g
percent purity = 1.96 g/2.5 g  X 100 %