What is the mass of sodium bromate solution necessary to prepare $85.5 c m^{3}$ of 0.672 N solution when the half-cell reaction is
$B r O_{3}^{-} + 6 H^{+} + 6 e^{-} \to B r^{-} + 3 H_{2} O$

1.146 g

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1.246 g

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1.346 g

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1.446 g

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Solution

The correct option is D 1.446 g
The half-cell reaction is
$B r O_{3}^{-} + 6 H^{+} + 6 e^{-} \to B r^{-} + 3 H_{2} O$
Since $6 e^{-}$ are involved in the reaction, we have
$= \frac{Normality}{6} = \frac{0.675}{6} = 0.112 M$
$Molar mass of N a B r O_{3} = 151 g m o l^{- 1}$
Mass of $N a B r O_{3}$ needed to prepare required solution
$= (\frac{0.112}{1000} \times 85.5) \times 151 = 1.446 g$
Hence option (d) is correct.

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Similar questions

85.5

0.672 N

B r O_{3}^{⊝} + 6 H^{⨁} + 6 e^{-} \to B r^{⊝} + 3 H_{2} O

B r O_{3}^{-} + 6 H^{+} + 6 e^{-} \to B r^{-} + 3 H_{2} O

85.4 m L

0.672 N

{B r O_{3}}^{-} + 6 H^{+} + 6 e^{-} \to B r^{-} + 3 H_{2} O

2 {B r O_{3}}^{-} + 12 H^{+} + 10 e^{-} \to B r_{2} + 6 H_{2} O

B r O_{3}^{-} + 6 H^{+} + 6 e^{-} \to B r^{-} + 3 H_{2} O

N a B r O_{3} + 6 H^{+} + 6 e^{-} \to N a B r + 3 H_{2} O

(N a B r O_{3})

0.6 N

100 m L

N a B r O_{3} = 151 g m o l^{- 1}

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