In a titration experiment, 10   mL of an FeCl 2 solution consumed 25   mL of a standard K 2 Cr 2 O 7 solution to reach the equivalent point. The standard K 2 Cr 2 O 7 solution is prepared by dissolving 1 . 225   g of K 2 Cr 2 O 7 in 250   mL water. The concentration of the FeCl 2 solution is closest to [Given : molecular weight of K 2 Cr 2 O 7 = 294   g   mol - 1 ]

Step-by-step solution

In titration at equivalence point, number of equivalents of both reactants must be same. ∴   N 1 V 1 = N 2 V 2 N (normality) of FeCl 2 = N × V   of   K 2 Cr 2 O 7 V   of   FeCl 2 N (normality) of K 2 Cr 2 O 7 =   Mass   × 1000   Equivalent   mass   ×   Volume   of   solution   prepared Equivalent mass =   Molar   mass   n   factor   = 294 6 = 49   g / equi . ( K 2 Cr 2 O 7 as oxidising agent is r