You visited us 1 times! Enjoying our articles? Unlock Full Access!

Assigning Oxidation Numbers

For the react...

Question

For the reaction, $F e_{0.95} O$ (molar mass M) $\to F e_{2} O_{3}$ , what is the equivalent weight of $F e_{0.95} O$ ?

M/0.85

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

M/0.95

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

M/0.80

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

M/0.90

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

Open in App

Solution

The correct option is A M/0.85
n-factor $= (3 - \frac{2}{0.95}) \times 0.95 = 0.85.$
Equivalent weight $= \frac{M o l e c u l a r w e i g h t / A t o m i c w e i g h t}{n_{f a c t o r}}$
$= M / 0.85$

.

Suggest Corrections

Similar questions

F e_{0.95} O

\to F e_{2} O_{3}

F e_{0.95} O

{F e}_{0.95} O

⟶ {F e}_{2} O_{3}

{F e}_{0.95} O

F e_{0.95} O (m o l a r m a s s : M) \to F e_{2} O_{3}

F e_{0.95} O

F e 0.95)

\to F e_{2} O_{3}

F e_{0.95} O

F e_{0.95} O

F e_{2} O_{3}

Join BYJU'S Learning Program