A Current of 9.65 ampere flowing for 10 minutes deposits 3.0g of the metal which is monovalent. The atomci 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 1: Calculate the total charge (Q) passed through the solution.
The formula to calculate charge is:
\[ Q = I \times t \]
Where:
- \( I \) = current in amperes (A)
- \( t \) = time in seconds (s)

Given:
- \( I = 9.65 \, \text{A} \)
- \( t = 10 \, \text{minutes} = 10 \times 60 = 600 \, \text{s} \)

Now, substituting the values:
\[ Q = 9.65 \, \text{A} \times 600 \, \text{s} = 5790 \, \text{C} \]

### Step 2: Determine the number of moles of electrons transferred.
According to Faraday's law, 1 mole of electrons corresponds to 96500 coulombs of charge. Thus, we can find the number of moles of electrons (n) using the formula:
\[ n = \frac{Q}{F} \]
Where:
- \( F = 96500 \, \text{C/mol} \) (Faraday's constant)

Substituting the values:
\[ n = \frac{5790 \, \text{C}}{96500 \, \text{C/mol}} \approx 0.060 \, \text{mol} \]

### Step 3: Relate moles of electrons to moles of metal deposited.
Since the metal is monovalent, it means that 1 mole of electrons will deposit 1 mole of the metal. Therefore, the moles of the metal deposited will also be:
\[ n_{\text{metal}} = 0.060 \, \text{mol} \]

### Step 4: Calculate the atomic mass of the metal.
We know that the mass of the metal deposited (m) is given as 3.0 g. The relationship between mass, moles, and molar mass (M) is given by:
\[ M = \frac{m}{n} \]
Where:
- \( m = 3.0 \, \text{g} \)
- \( n = 0.060 \, \text{mol} \)

Substituting the values:
\[ M = \frac{3.0 \, \text{g}}{0.060 \, \text{mol}} = 50 \, \text{g/mol} \]

### Conclusion:
The atomic mass of the monovalent metal is **50 g/mol**.

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