The carbonate of a metal is isomorphous with `MgCO_(3)` and contains 10.34 percent of carbon. Atomic mass of metal will be:

Options

A. 56
B. 98
C. 40
D. 23

Answer

To find the atomic mass of the metal in the carbonate that is isomorphous with MgCO₃ and contains 10.34% carbon, we can follow these steps:

### Step 1: Understand the composition of the metal carbonate
Since the carbonate is isomorphous with MgCO₃, we can represent the metal carbonate as MCO₃, where M is the metal. The molecular weight of MCO₃ can be expressed as:
\[ \text{Molecular weight of MCO}_3 = x + 12 + 3 \times 16 \]
Where:
- \( x \) = atomic mass of the metal
- 12 = atomic mass of carbon (C)
- 16 = atomic mass of oxygen (O), and there are 3 oxygen atoms.

Calculating the weight of the oxygen:
\[ 3 \times 16 = 48 \]
Thus, the molecular weight of MCO₃ becomes:
\[ \text{Molecular weight of MCO}_3 = x + 12 + 48 = x + 60 \]

### Step 2: Set up the equation based on the percentage of carbon
According to the problem, the percentage of carbon in the metal carbonate is given as 10.34%. We can express this mathematically as:
\[ \frac{\text{Weight of Carbon}}{\text{Molecular Weight of MCO}_3} \times 100 = 10.34 \]
The weight of carbon in MCO₃ is 12 grams. Therefore, we can write:
\[ \frac{12}{x + 60} \times 100 = 10.34 \]

### Step 3: Solve the equation for x
Rearranging the equation gives:
\[ 12 \times 100 = 10.34 \times (x + 60) \]
\[ 1200 = 10.34x + 620.4 \]
Subtracting 620.4 from both sides:
\[ 1200 - 620.4 = 10.34x \]
\[ 579.6 = 10.34x \]
Now, divide both sides by 10.34:
\[ x = \frac{579.6}{10.34} \]
Calculating this gives:
\[ x \approx 56.05 \]

### Step 4: Conclusion
Thus, the atomic mass of the metal is approximately 56.