12 gm of an alkaline earth metal gives 14.8 g of its nitride. The atomic mass of metal is

Options

A. 12
B. 24
C. 20
D. 40

Explanation

To find the atomic mass of the alkaline earth metal that gives 14.8 g of its nitride from 12 g of the metal, we can follow these steps:

### Step 1: Identify the reaction and the formula of the nitride
Alkaline earth metals typically form compounds in the +2 oxidation state. The nitride of nitrogen is represented as \(N_2^{2-}\). Therefore, the formula for the nitride formed with the alkaline earth metal (M) can be written as:
\[ M_3N_2 \]

### Step 2: Use the conservation of mass
According to the law of conservation of mass, the total mass of the reactants must equal the total mass of the products. We have:
- Mass of alkaline earth metal (M) = 12 g
- Mass of nitride (M3N2) = 14.8 g

To find the mass of nitrogen in the nitride, we can use:
\[ \text{Mass of nitrogen} = \text{Mass of nitride} - \text{Mass of metal} \]
\[ \text{Mass of nitrogen} = 14.8 \, \text{g} - 12 \, \text{g} = 2.8 \, \text{g} \]

### Step 3: Calculate moles of nitrogen
The molar mass of nitrogen (\(N_2\)) is 28 g/mol. To find the number of moles of nitrogen:
\[ \text{Number of moles of } N_2 = \frac{\text{Mass of nitrogen}}{\text{Molar mass of } N_2} = \frac{2.8 \, \text{g}}{28 \, \text{g/mol}} = 0.1 \, \text{mol} \]

### Step 4: Relate moles of nitrogen to moles of metal
From the formula \(M_3N_2\), we see that 1 mole of \(N_2\) reacts with 3 moles of metal (M). Therefore, if we have 0.1 moles of \(N_2\), the moles of metal will be:
\[ \text{Moles of metal} = 3 \times \text{Moles of } N_2 = 3 \times 0.1 = 0.3 \, \text{mol} \]

### Step 5: Calculate the atomic mass of the metal
The atomic mass (M) can be calculated using the formula:
\[ \text{Atomic mass} = \frac{\text{Mass of metal}}{\text{Number of moles of metal}} = \frac{12 \, \text{g}}{0.3 \, \text{mol}} = 40 \, \text{g/mol} \]

### Final Answer
The atomic mass of the alkaline earth metal is **40 g/mol**.

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