Calculate the mass of Magnesium that will be liberated from its salt by the same quantity of electricity that liberated 16.0 g of Silver. [Mg = 24.0, Ag = 108 ]

To solve this problem, we use Faraday’s Second Law of Electrolysis, which states that when the same quantity of electricity is passed through different electrolytes, the masses of the substances deposited are proportional to their chemical equivalents (molar mass divided by valency).

1. Analyze the question

• We need to find the mass of Magnesium ($Mg$) deposited.
• We are given the mass of Silver ($Ag$) deposited = 16.0 g.
• Molar mass of $Mg = 24.0\text{ g/mol}$.
• Molar mass of $Ag = 108.0\text{ g/mol}$.
• Valency of $Mg = 2$ (since it forms $Mg^{2+}$ ions).
• Valency of $Ag = 1$ (since it forms $Ag^+$ ions).

2. Explain the concept

The relationship is given by the formula:

Where Equivalent mass = $\frac{\text{Molar mass}}{\text{Valency}}$.

3. Show the solution

Step 1: Calculate the equivalent masses.

• Equivalent mass of $Ag = \frac{108}{1} = 108$
• Equivalent mass of $Mg = \frac{24}{2} = 12$

Step 2: Set up the ratio. Let $x$ be the mass of Magnesium.

Step 3: Solve for $x$.

Rounding to two decimal places, we get 1.78 g.

4. Identify the answer

Comparing our result to the options provided, the correct value is 1.78 g. This matches option D.

5. Exam Tip

Always remember to check the valency of the metals involved! A common mistake students make is using the molar mass directly without dividing by the valency (the charge of the ion). For Group 1 metals (like $Ag, Na, K$), the valency is 1; for Group 2 metals (like $Mg, Ca$), the valency is 2.