A current of `9.65A` flowing for `10 mi n` deposits `3.0g` of the metal which is monovalent. The atomic mass of the metal is

Options

A. 10
B. 50
C. 30
D. `96.5`

Explanation

To find the atomic mass of the monovalent metal deposited by the current, we can follow these steps:

### Step-by-Step Solution

1. **Convert Time to Seconds**:
The current is given for 10 minutes. We need to convert this time into seconds.
\[ \text{Time in seconds} = 10 \text{ minutes} \times 60 \text{ seconds/minute} = 600 \text{ seconds} \]

2. **Calculate Total Charge (Q)**:
Using the formula \( Q = I \times t \), where \( I \) is the current in amperes and \( t \) is the time in seconds:
\[ Q = 9.65 \text{ A} \times 600 \text{ s} = 5790 \text{ C} \]

3. **Convert Charge to Moles of Electrons**:
We know that 1 mole of electrons corresponds to 96500 coulombs (Faraday's constant). Therefore, the number of moles of electrons (n) can be calculated as:
\[ n = \frac{Q}{F} = \frac{5790 \text{ C}}{96500 \text{ C/mol}} \approx 0.05996 \text{ mol} \]

4. **Relate Moles of Electrons to Moles of Metal**:
Since the metal is monovalent, 1 mole of electrons will deposit 1 mole of the metal. Thus, the moles of metal deposited are also approximately 0.05996 mol.

5. **Calculate Mass of Metal Deposited**:
We are given that the mass of the metal deposited is 3.0 g. We can use this to find the atomic mass (M) of the metal:
\[ \text{Mass} = \text{Moles} \times \text{Molar Mass} \implies M = \frac{\text{Mass}}{\text{Moles}} = \frac{3.0 \text{ g}}{0.05996 \text{ mol}} \approx 50.01 \text{ g/mol} \]

6. **Conclusion**:
The atomic mass of the metal is approximately 50 g/mol. Therefore, the answer is:
\[ \text{Atomic Mass} = 50 \text{ g/mol} \]

### Final Answer
The atomic mass of the metal is **50 g/mol**.

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