Question

❤❤❤❤❤❤❤❤❤❤❤❤
.
.
.
.
.
.
❤amount of FeSO4 , solution required containing 20% (w/w) FeSO4 for complete titration of 100 ml of 0.1M KMnO4 solution in acidic medium is
A→7.6g
B→38 g
C→76g
D →3.8g
.
.
.
i will mark as brainlist.... please no spam ​

Answer 1

Answer:

20.8 g of FeSO4

Explanation:

weight of FeSO4 - ?

mass of FeSO4 - 151.9 g/mol

weight of KMnO4 - 20g

mass of KMnO4 - 158 g/mol

Molarity of KMnO4 - 0.1 M

M = w of FeSO4 × m of FeSO4 + w of KMnO4 × m of KMnO4

0.1 = w of FeSO4 × 151.9 + 20 × 158

0.1 = w of FeSO4 × 151.9 + 3160

3159.9 = w of FeSO4 × 151.9

3159.9/ 151.9 = w of FeSO4

20.8 = w of FeSO4

hence proved