\(30\, \ mL\) of \(NiSO_4\) is electrolysed using a current of \(360\, \ mA\) for \(30\) minutes. How much of the metal would have been plated out if the current efficiency is only \(60\%\)? (\(Ni = 58.7\, \ g/mol)\)
Answer & explanation
Correct answer: option 3
The correct answer is option 3. 0.1182 g.
The amount of substance deposited:
\[ \text{Amount of substance} = \frac{648\, \text{C}}{2 \times 96500\, \text{C/mol}} = \frac{648}{193000}\, \text{mol} \]
Now, considering a current efficiency of \(60\%\), only \(60\%\) of this substance will be plated out. So, the actual amount of metal plated out is:
\[ \text{Amount of metal plated out} = 0.60 \times \frac{648}{193000}\, \text{mol} \]
Now, let's calculate this value:
\[ \text{Amount of metal plated out} = 0.60 \times \frac{648}{193000} = \frac{0.60 \times 648}{193000} \]
\[ \text{Amount of metal plated out} = 0.00201450777 \, \text{mol} \]
Now, we can calculate the mass of the metal using its molar mass:
\[ \text{Mass of metal} = \text{Amount of metal plated out} \times \text{Molar mass of Ni} \]
Given that the molar mass of Ni is \( 58.7\, \text{g/mol} \), we have:
\[ \text{Mass of metal} = 0.00201450777 \times 58.7\, \text{g} \]
\[ \text{Mass of metal} = 0.1182\, \text{g} \]
So, the correct answer is option (3) \(0.1182\, \text{g}\)