The sulphate of a metal M contains 20% of M. This sulphate is isomorphous with `ZnSO_(4).7H_(2)O`. The atomic mass of M is

Options

A. `12`
B. `24`
C. `36`
D. `48`

Explanation

To find the atomic mass of the metal M in the sulfate, we can follow these steps:

### Step 1: Understand the composition of the sulfate
We know that the sulfate of metal M is isomorphous with `ZnSO4.7H2O`. This means it has a similar structure and formula, which can be represented as `MSO4.7H2O`.

### Step 2: Determine the total mass of the compound
The total mass of the compound `MSO4.7H2O` can be calculated as follows:
- Atomic mass of M = A
- Atomic mass of sulfur (S) = 32 g/mol
- Atomic mass of oxygen (O) = 16 g/mol
- The mass of sulfate (SO4) = 32 + (4 × 16) = 32 + 64 = 96 g/mol
- The mass of 7 water molecules (7H2O) = 7 × (2 + 16) = 7 × 18 = 126 g/mol

So, the total mass of the compound is:
\[ \text{Total mass} = A + 96 + 126 = A + 222 \]

### Step 3: Set up the equation based on the percentage composition
We know that the metal M constitutes 20% of the total mass of the compound. Therefore, we can set up the equation:
\[ \frac{A}{A + 222} = 0.20 \]

### Step 4: Solve the equation
To eliminate the fraction, we can cross-multiply:
\[ A = 0.20(A + 222) \]

Expanding this gives:
\[ A = 0.20A + 44.4 \]

Now, rearranging the equation:
\[ A - 0.20A = 44.4 \]
\[ 0.80A = 44.4 \]

Dividing both sides by 0.80:
\[ A = \frac{44.4}{0.80} \]
\[ A = 55.5 \]

### Conclusion
The atomic mass of metal M is **55.5 g/mol**.

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