You visited us 2 times! Enjoying our articles? Unlock Full Access!

Equivalent Mass

Find the equi...

Question

Find the equivalent weight of $K M n O_{4}$ in the given reaction, if the molar mass of $M n O_{4}^{-}$ is 'M' ?
$K M n O_{4} \to M n^{2 +}$ (acidic medium)

41.6

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

21.6

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

11.6

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

31.6

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

Open in App

Solution

The correct option is D $31.6$
In acidic medium
$(+ 7) K M n O_{4} \to (+ 2) M n^{2 +}$
$n_{f} = | 2 - 7 | \times 1 = 5$
Here 5 electrons are taken so
$equivalent weight = \frac{M o l . w t . o f K M n O_{4}}{5} = \frac{158}{5} = 31.6$

Suggest Corrections

Similar questions

K M n O_{4}

M n O_{4}^{-}

K M n O_{4} \to M n^{2 +}

K M n O_{4}

M n O_{4}^{-}

M n O_{4}^{-} \to M n^{4 +}

K M n O_{4} + 8 H^{+} + 5 e^{-} \to K^{+} + M n^{2 +} + 4 H_{2} O

K M n O_{4}

K M n O_{4} = 158

K M n O_{4}

31.6

M n O_{4}^{-}

M n^{2 +}

K M n O_{4}

M n O_{4}^{-} (+ 7) \to M n O_{4}^{- 2} (+ 6)

Join BYJU'S Learning Program

Explore more

Join BYJU'S Learning Program