Question

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 :

• A. 64
• B. 40
• C. 24
• D. 12

Answer 1

Answer: (B)

The quantity of electricity passed $=Q\left(C\right)=I\left(A\right)\times t\left(s\right)=3.86A\times 3000s=11580C$.
The gram atomic weight of divalent metal = W $\frac{g}{mol}$
Moles of electrons passed $=\frac{Q\left(C\right)}{96500\frac{C}{mol{e}^{-}}}=\frac{11580C}{96500\frac{C}{mol{e}^{-}}}=\frac{11580}{96500}mol{e}^{-}$
The mass of divalent metal deposited = 2.4 g $=atomicweightofmetal\times moleratio\times molesofelectronspassed=W\frac{g}{mol}\times \frac{1molmetal}{2mol{e}^{-}}\times \frac{11580}{96500}mol{e}^{-}W=40\frac{g}{mol}.$