1 g of arsenic dissolved in 86 g of benzene brings down the freezing point to `5.31^@C` from `5.50^@C`.If `K_f` of benzene is `4.9 (.^(@)C)/m`, the atomicity of the molecule is : (As =75 or As GAM 75)

Explanation

To solve the problem, we will follow these steps:

### Step 1: Write down the given data
- Mass of arsenic (Wv) = 1 g
- Molar mass of arsenic (Mb) = 75 g/mol
- Mass of benzene (Wa) = 86 g
- Freezing point depression (ΔTf) = 5.50°C - 5.31°C = 0.19°C
- Freezing point depression constant (Kf) of benzene = 4.9 °C/m

### Step 2: Use the formula for freezing point depression
The formula for freezing point depression is given by:
\[ \Delta T_f = i \cdot K_f \cdot m \]
Where:
- \(i\) = van 't Hoff factor (which indicates the number of particles the solute breaks into)
- \(m\) = molality of the solution

### Step 3: Calculate the molality (m)
Molality (m) is defined as:
\[ m = \frac{Wv}{Wa} \cdot \frac{1000}{Mb} \]
Substituting the values:
\[ m = \frac{1 \, \text{g}}{86 \, \text{g}} \cdot \frac{1000}{75 \, \text{g/mol}} = \frac{1}{86} \cdot \frac{1000}{75} \]
Calculating this gives:
\[ m = \frac{1000}{6450} \approx 0.155 \, \text{mol/kg} \]

### Step 4: Substitute values into the freezing point depression formula
Now substituting the values into the freezing point depression formula:
\[ 0.19 = i \cdot 4.9 \cdot 0.155 \]
Rearranging to solve for \(i\):
\[ i = \frac{0.19}{4.9 \cdot 0.155} \]
Calculating this gives:
\[ i \approx \frac{0.19}{0.7595} \approx 0.25 \]

### Step 5: Determine atomicity
The van 't Hoff factor \(i\) represents the number of particles the solute dissociates into. If \(i = 0.25\), it indicates that the molecule of arsenic (As) is likely to be tetratomic (4 atoms).

### Conclusion
Thus, the atomicity of the arsenic molecule is 4.