40 gm of a carbonate of an alkali metal or alkaline earth metal containing some insert impurities was made to react with excess HCl solution. The liberated `CO_(2)` occupied 12.315 lit. at 1 atm & 300 K . The correct option is

Options

A. Mass of impurity is 1 gm and metal is Be
B. Mass of impurity is 3 gm and metal is Li
C. Mass of impurity is 5 gm and metal is Be
D. Mass of impurity is 2 gm and metal is Mg

Answer

To solve the problem step by step, let's break it down:

### Step 1: Calculate the moles of CO2 produced
We know that the volume of CO2 liberated is 12.315 liters at 1 atm and 300 K. We can use the ideal gas law to find the number of moles of CO2.

Using the ideal gas equation:
\[ PV = nRT \]

Where:
- \( P = 1 \, \text{atm} \)
- \( V = 12.315 \, \text{liters} \)
- \( R = 0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1} \) (universal gas constant)
- \( T = 300 \, \text{K} \)

Rearranging the equation to solve for \( n \):
\[ n = \frac{PV}{RT} \]

Substituting the values:
\[ n = \frac{(1 \, \text{atm})(12.315 \, \text{L})}{(0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1})(300 \, \text{K})} \]

Calculating:
\[ n = \frac{12.315}{24.63} \approx 0.5 \, \text{moles of CO2} \]

### Step 2: Determine the moles of the carbonate
From the stoichiometry of the reaction, we know that 1 mole of a carbonate (MCO3) produces 1 mole of CO2. Therefore, if 0.5 moles of CO2 are produced, then 0.5 moles of the carbonate (MCO3) were present.

### Step 3: Calculate the mass of the carbonate
To find the mass of the carbonate, we need its molar mass. The molar mass of the carbonate (MCO3) can be calculated as follows:
- Molar mass of CO3 = 12 (C) + 3 × 16 (O) = 60 g/mol
- Let the molar mass of M be the molar mass of the alkali or alkaline earth metal.

Thus, the molar mass of MCO3 is:
\[ \text{Molar mass of MCO3} = \text{M} + 60 \]

The mass of the carbonate can be calculated using:
\[ \text{mass} = \text{moles} \times \text{molar mass} \]
\[ 40 \, \text{g} = 0.5 \times (\text{M} + 60) \]

### Step 4: Solve for M
Rearranging the equation:
\[ 40 = 0.5 \times (\text{M} + 60) \]
\[ 80 = \text{M} + 60 \]
\[ \text{M} = 80 - 60 = 20 \, \text{g/mol} \]

### Step 5: Identify the metal
The molar mass of 20 g/mol corresponds to lithium (Li), which has a molar mass of approximately 6.94 g/mol. The carbonate would be lithium carbonate (Li2CO3).

### Step 6: Calculate the mass of impurities
The mass of the carbonate (Li2CO3) can be calculated:
\[ \text{Molar mass of Li2CO3} = 2 \times 6.94 + 12 + 3 \times 16 = 73.89 \, \text{g/mol} \]
\[ \text{Mass of Li2CO3} = 0.5 \, \text{moles} \times 73.89 \, \text{g/mol} = 36.945 \, \text{g} \]

Thus, the mass of impurities is:
\[ \text{Impurities} = 40 \, \text{g} - 36.945 \, \text{g} \approx 3.055 \, \text{g} \]

### Conclusion
The mass of impurities is approximately 3 g, which matches the option given in the problem.

### Final Answer
The correct option is that the mass of impurities is approximately 3 g.

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