What is the equivalent mass of $I O_{4}^{-}$ when it is converted into $I_{2}$ in acid medium

\frac{M}{6}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{M}{7}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{M}{5}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{M}{4}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

None of these

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $\frac{M}{7}$
Equivalent mass
$= \frac{M o l e c u l a r w e i g h t}{C h a n g e i n o x i d a t i o n n u m b e r p e r m o l e}$
Suppose molecular weight is M
Oxidation number of $I_{2} i n I O_{4}^{-}$ in
Acidic medium i.e., $I \times (- 8) + I e^{-} = + 7$
So eq. wt. $\frac{M}{7}$ .

Suggest Corrections

Similar questions

I O_{4}^{-}

I_{2}

I O_{4}^{-}

I_{2}

M n S O_{4}

M

K M n O_{4}

K_{2} M n O_{4}

N_{2} H_{4}

I O_{3}^{-}

I^{+}

N_{2} H_{4} + I O_{3}^{-} + H^{+} ⟶ N_{2} + I^{+} + H_{2} O

N_{2} H_{4}

^{- 1}

Join BYJU'S Learning Program