$2 C u S O_{4} - ------- - + 4 K I ⟶ {C u}_{2} I_{2} + 2 K_{2} S O_{4} + I_{2}$
The equivalent weight of the underlined compound by taking its molecular weight as $M$ is :

E = M

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

E = \frac{M}{3}

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

E = \frac{M}{2}

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 A $E = M$
The oxidation number of Cu in $C u S O_{4}$ is 2.
The oxidation number of Cu in $C u_{2} I_{2}$ is 1.
Thus, the oxidation number of Cu changes by 1.
$Equivalent weight = \frac{Molecular weight}{Change in the oxidation number}$
Hence, $E = \frac{M}{1}$

Suggest Corrections

Similar questions

P - - {C l}_{5} ⟶ P {C l}_{3} + {C l}_{2}

M

2 C u S O_{4} + 4 K I \to C u_{2} I_{2} + 2 K_{2} S O_{4} + I_{2}

C u S O_{4}

(M w = 159.5 g

m o l^{- 1})

2 {H S}^{-} - ----- - + 4 H S O_{3}^{-} ⟶ 3 S_{2} O_{3}^{2 -} + 3 H_{2} O

M

M

N H_{4} {N O}_{3} - --------- - ⟶ N_{2} O + 2 H_{2} O

4 {F e}^{2 +} + 2 {N H}_{2} O H - --------- - ⟶ 4 {F e}^{3 +} + N_{2} O + 4 H^{+}

M

Join BYJU'S Learning Program