The weight of sodium bromate required to prepare 85.5 ml of 0.672 N solution for cell reaction BrO3– + 6H+ + 6e → Br– + 3H2O is

Q: The weight of sodium bromate required to prepare 85.5 ml of 0.672 N solution for cell reaction BrO3 + 6H+ + 6e → Br + 3H2O is

(A) 1.56 gm

(B) 1.45 gm

(C) 1.23 gm

(D) 1.32 gm

Sol: Meq. of NaBrO3 = 85.5 × 0.672 = 57.456

Let weight of NaBrO3 = W

$\large \frac{W}{M_{NaBrO_3}}\times 6 \times 1000= 57.456 $ (equivalent weight = M/6) of n-factor = 6

$\large \frac{W}{151}\times 6 \times 1000= 57.456 $

W = 1.45 gm

Ans: (B)