Excess of milliequivalents of $B r O_{3}^{⊝} =$ milliequivalents of $A s O_{2}^{⊝}$ in reaction.

True

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False

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Solution

The correct option is A True
In the second equation, excess $B r O_{3}^{-}$ completely reacts with $A s O_{2}^{-}$
Hence, the excess of meq. of $B r O_{3}^{-}$ is equal to the meq. of $A s O_{2}^{-}$ .
Hence, the given statement is true.

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