One gm metal `M^(3+)` was discharged by the passage of `1.81xx10^(23)` electrons.What is the atomic mass of metal?

Options

A. (a) 33.35
B. (b) 133.4
C. (c) 66.7
D. (d) None of these

Answer

To solve the problem, we need to find the atomic mass of the metal \( M^{3+} \) that was discharged by the passage of \( 1.81 \times 10^{23} \) electrons. We will use Faraday's laws of electrolysis to derive the atomic mass.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Weight of the metal \( M^{3+} \) = 1 g
- Number of electrons \( n = 1.81 \times 10^{23} \)
- Charge of one electron \( e = 1.6 \times 10^{-19} \) C

2. **Calculate the Total Charge (Q):**
The total charge \( Q \) can be calculated using the formula:
\[ Q = n \times e \]
Substituting the values:
\[ Q = (1.81 \times 10^{23}) \times (1.6 \times 10^{-19}) = 2.896 \times 10^{4} \text{ C} \]

3. **Determine the n-factor:**
Since the metal is \( M^{3+} \), the n-factor (number of electrons transferred per ion) is 3.

4. **Use Faraday's First Law of Electrolysis:**
According to Faraday's first law:
\[ \text{Weight} = \frac{Z \times Q}{96500} \]
where \( Z \) is the electrochemical equivalent. We can express \( Z \) in terms of molar mass \( M \) and n-factor:
\[ Z = \frac{M}{n} \]
Therefore, we can rewrite the equation as:
\[ \text{Weight} = \frac{M}{n} \times \frac{Q}{96500} \]

5. **Substituting the Known Values:**
We know the weight of the metal is 1 g, and substituting \( n = 3 \):
\[ 1 = \frac{M}{3} \times \frac{2.896 \times 10^{4}}{96500} \]

6. **Rearranging the Equation to Solve for M:**
Rearranging gives:
\[ M = \frac{1 \times 3 \times 96500}{2.896 \times 10^{4}} \]

7. **Calculating M:**
\[ M = \frac{289500}{28960} \approx 10 \text{ g/mol} \]

8. **Conclusion:**
The atomic mass of the metal \( M \) is approximately 10 g/mol.