UPSEE Solved Model Paper 7

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