Question

Volume of 0.1 M K,Cr, required to oxidise
35 ml of 0.5 M FeSO, solution is -
(A) 29 ml
(C) 175 ml
(B) 87 ml
(D) 145 ml​

Answer 1

Answer:

(A) 29 ml

The chemical equation is:

K2Cr2O7 + 6FeSO4 + 7H2SO4 → Cr2(SO4)3 + 3Fe2(SO4)3 + K2SO4 + 7H2O

Moles of FeSO4 in 35ml of 0.5M FeSO4 solution = (35/1000) * 0.5 = 0.0175

According to the equation:

6 moles of FeSO4 require 1 mole of K2Cr2O7

0.0175 moles will require = (0.0175x1)/6

                                            = 0.002917 moles of K2Cr2O7

V=  Number of moles/Molarity

V of K2Cr2O7 = 0.002917/0.1 = 0.02917 L

V = 0.02917 x 1000 ml

 V= 29.17 ml