$B r O_{3}^{-}$ is oxidised to $B r O_{4}^{-}$ by $X e F_{2}$

$X e F_{2} m_{1} + B r O_{3}^{-} m_{2} + H_{2} O ⟶ B r O_{4}^{-} + X e + 2 H F$
where,
$m_{1}$ and $m_{2}$ are the molar masses of $X e F_{2}$ and $B r O_{3}^{-}$ , respectively.

Ratio of equivalent mass of $X e F_{2}$ and $B r O_{3}^{-}$ is_________.

\frac{m_{1}}{m_{2}}

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\frac{2 m_{1}}{m_{2}}

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\frac{m_{1}}{2 m_{2}}

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\frac{m_{1}}{3 m_{2}}

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Solution

The correct option is B $\frac{m_{1}}{m_{2}}$
$X e F_{2} m_{1} + B r O_{3}^{-} m_{2} + H_{2} O ⟶ B r O_{4}^{-} + X e + 2 H F$
molar mass
+2 +5 +7 0
So eq wt. of $X e F_{2} = m o l a r m a s s / 2$
So eq wt. of $B r O_{3}^{-} = m o l a r m a s s / 2$
Therefore their eq. wt. ratio is $m_{1} / m_{2}$

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Similar questions

Q. In the following reaction,

X e F_{2} + B r O_{3}^{-} + H_{2} O ⟶ B r O_{4}^{-} + X e + 2 H F

In the following reaction

X e F_{2} + {B r O_{3}}^{-} + H_{2} O ⟶ X e + {B r O_{4}}^{-} + 2 H F

e q u i v a l e n t m a s s o f {B r O_{3}}^{-} = \frac{m o l a r m a s s}{. . . . . . . . . .}

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BrO−3 is oxidised to BrO−4 by XeF2 XeF2m1+BrO−3m2+H2O⟶BrO−4+Xe+2HFwhere, m1 and m2 are the molar masses of XeF2 and BrO−3, respectively.Ratio of equivalent mass of XeF2 and BrO−3 is_________.

The correct option is B m1m2XeF2m1+BrO−3m2+H2O⟶BrO−4+Xe+2HFmolar mass +2 +5 +7 0So eq wt. of XeF2=molarmass/2 So eq wt. of BrO−3=molarmass/2Therefore their eq. wt. ratio is m1/m2