When 3.86 amperes current is passes through an electrolyte for 50 minutes, 2.4 grams of a divalent metal is deposited. The gram atomic weight of the metal (in grams) is :

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Solution

The correct option is B 40
The quantity of electricity passed $= Q (C) = I (A) \times t (s) = 3.86 A \times 3000 s = 11580 C$ .
The gram atomic weight of divalent metal = W $\frac{g}{m o l}$
Moles of electrons passed $= \frac{Q (C)}{96500 \frac{C}{m o l e^{-}}} = \frac{11580 C}{96500 \frac{C}{m o l e^{-}}} = \frac{11580}{96500} m o l e^{-}$
The mass of divalent metal deposited = 2.4 g $= a t o m i c w e i g h t o f m e t a l \times m o l e r a t i o \times m o l e s o f e l e c t r o n s p a s s e d = W \frac{g}{m o l} \times \frac{1 m o l m e t a l}{2 m o l e^{-}} \times \frac{11580}{96500} m o l e^{-} W = 40 \frac{g}{m o l} .$

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