The sulphate of a metal `M` contains `9.87%` of `M`, This sulphate is isomorphous with `ZnSO_(4).7H_(2)O`. What is the atomic weight of `M` ?

Options

A. 40.3
B. 36.3
C. 24.3
D. 11.3

Answer

To find the atomic weight of the metal \( M \) in the sulfate \( MSO_4 \cdot 7H_2O \) that contains \( 9.87\% \) of \( M \), we can follow these steps:

### Step 1: Write the formula for the sulfate
Since the sulfate of metal \( M \) is isomorphous with \( ZnSO_4 \cdot 7H_2O \), we can write the formula as:
\[ MSO_4 \cdot 7H_2O \]

### Step 2: Calculate the molar mass of the compound
The molar mass of \( MSO_4 \cdot 7H_2O \) can be calculated as follows:
- Atomic mass of \( M \) = \( M \)
- Atomic mass of sulfur (S) = \( 32 \, \text{g/mol} \)
- Atomic mass of oxygen (O) = \( 16 \, \text{g/mol} \) (4 oxygen atoms in sulfate)
- Molar mass of water (H₂O) = \( 18 \, \text{g/mol} \) (7 water molecules)

Now, calculate the total molar mass:
\[ \text{Total molar mass} = M + 32 + (16 \times 4) + (7 \times 18) \]
\[ = M + 32 + 64 + 126 \]
\[ = M + 222 \]

### Step 3: Set up the percentage equation
Given that the sulfate contains \( 9.87\% \) of \( M \), we can set up the equation:
\[ \frac{M}{M + 222} \times 100 = 9.87 \]

### Step 4: Solve for \( M \)
First, simplify the equation:
\[ \frac{M}{M + 222} = \frac{9.87}{100} \]
Cross-multiply to eliminate the fraction:
\[ 100M = 9.87(M + 222) \]
Distributing \( 9.87 \):
\[ 100M = 9.87M + 2193.54 \]
Now, isolate \( M \):
\[ 100M - 9.87M = 2193.54 \]
\[ 90.13M = 2193.54 \]
Now, divide both sides by \( 90.13 \):
\[ M = \frac{2193.54}{90.13} \approx 24.3 \]

### Conclusion
The atomic weight of metal \( M \) is approximately \( 24.3 \, \text{g/mol} \).

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