The weight of sodium bromate required to prepare 55.5 mL of 0.672 N solution for cell reaction, BrO 3-+6 H ++6 e - arrow Br -+3 H 2 O , is

Question

The weight of sodium bromate required to prepare 55.5 mL of 0.672 N solution for cell reaction, BrO 3-+6 H ++6 e - arrow Br -+3 H 2 O , is (A) 1.56 g (B) 0.9386 g (C) 1.23 g (D) 1.32 g

Tardigrade Admin · Accepted Answer

Meq. of NaBrO 3=55.5 × 0.672=37.296 Let weight of NaBrO 3= W ∴ ( W / M NaBrO 3) × 6 × 1000=37.296 (equivalent weight = M/6) of n-factor = 6 ∴ ( M /151) × 6 × 1000=37.296 M = 0.9386 g

Q. The weight of sodium bromate required to prepare $55.5 m L$ of $0.672 N$ solution for cell reaction,
$B r O_{3}^{-} + 6 H^{+} + 6 e^{-} \to B r^{-} + 3 H_{2} O,$ is

1.56 g

0.9386 g

1.23 g

1.32 g

Meq. of $N a B r O_{3} = 55.5 \times 0.672 = 37.296$

Let weight of $N a B r O_{3} = W$

$∴ \frac{W}{M _{N a B r O_{3}}} \times 6 \times 1000 = 37.296$

(equivalent weight $= M /6)$ of $n$ -factor $= 6$

$∴ \frac{M}{151} \times 6 \times 1000 = 37.296$

$M = 0.9386 g$

Q. The weight of sodium bromate required to prepare 55.5mL of 0.672N solution for cell reaction, BrO3−​+6H++6e−→Br−+3H2​O, is