Given : Concentration of Na2CrO4 solution = 0.154 M Concentration of solution = 0 246 M Volume of N2S2O3 solution = 40 mL We have to find the value of Na2CrO4 solution used CrO42− → Cr(OH)4− Oxidation state of Cr in CrO42− = + 6 Oxidation state of Cr in Cr(OH)4− = + 3 Change in oxidation state, i.e. n-factor = 3 Similarly. S2O32− → SO42− Oxidation state of S in S2O32− = + 2 Oxidation state of S in S2O42− = + 6 Change in oxidation no. for one S atom = +6 - 2 = + 4 Change in oxidation no for two S atoms = 2 x (+4) = +8 Thus, n-factor is 8 Now On substituting the values in equation (i). w e get 0.154 × 3 × VNa2CrO4 = 0.246 × 8 × 40 VNa2CrO4 =
40×8×0.246
3×0.154
= 170.389 mL ~ 170.4 mL Hence, option 'D' is correct