1g mixture of equal number of mole of `Li_(2)CO_(3)` and other metal carbonate `(M_(2)CO_(3))` required 21.6mL of 0.5 N HCl for complete neutralisation reaction. What is the apoproximate atomic mass of the other metal?

Options

A. 25
B. 23
C. 51
D. 118

Answer

To find the approximate atomic mass of the other metal in the mixture of lithium carbonate (Li₂CO₃) and another metal carbonate (M₂CO₃), we can follow these steps:

### Step 1: Define the Variables
Let:
- \( x \) = mass of Li₂CO₃ in grams
- \( 1 - x \) = mass of M₂CO₃ in grams
- The total mass of the mixture = 1 g

### Step 2: Write the Equations for Moles
Since the number of moles of Li₂CO₃ is equal to the number of moles of M₂CO₃, we can express this as:
\[ \text{Number of moles of Li}_2\text{CO}_3 = \frac{x}{74} \]
\[ \text{Number of moles of M}_2\text{CO}_3 = \frac{1 - x}{(2m + 60)} \]
Where \( 74 \) is the molar mass of Li₂CO₃ and \( (2m + 60) \) is the molar mass of M₂CO₃.

### Step 3: Set Up the Equation for Neutralization
The mixture requires 21.6 mL of 0.5 N HCl for complete neutralization. The number of equivalents of HCl used can be calculated as:
\[ \text{Number of equivalents of HCl} = \text{Normality} \times \text{Volume (L)} = 0.5 \times \frac{21.6}{1000} = 0.0108 \text{ equivalents} \]

### Step 4: Write the Equation for Equivalents
The total number of equivalents from the carbonates is given by:
\[ \text{Number of equivalents from Li}_2\text{CO}_3 + \text{Number of equivalents from M}_2\text{CO}_3 = 0.0108 \]
Since both carbonates have a valency factor of 2:
\[ 2 \cdot \left(\frac{x}{74}\right) + 2 \cdot \left(\frac{1 - x}{(2m + 60)}\right) = 0.0108 \]

### Step 5: Simplify the Equation
This simplifies to:
\[ \frac{2x}{74} + \frac{2(1 - x)}{(2m + 60)} = 0.0108 \]

### Step 6: Substitute \( x \) in Terms of \( m \)
From the first equation, we can express \( x \) in terms of \( m \):
\[ x = \frac{2(1 - x)}{(2m + 60)} \cdot 74 \]

### Step 7: Solve for \( m \)
Substituting \( x \) back into the equation and solving for \( m \):
1. Rearrange and solve for \( m \):
\[ 2x + 2(1 - x) = 0.0108(2m + 60) \]
2. Substitute \( x = 1 - (1 - x) \) and solve for \( m \).

### Step 8: Calculate the Atomic Mass
After solving the equations, we find that:
\[ m \approx 118 \text{ g/mol} \]

### Conclusion
The approximate atomic mass of the other metal (M) is **118 g/mol**.

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