`2.56g` of sulfur in `100 g` of `CS_(2)` of has depression in freez point of `0.01^(@)C.K_(f)=0.1^(@)` mol `al^(-1)`. Hence, he atomicity of sulfur is `CS_(2)` is

Explanation

To solve the problem step by step, we will follow the given information and apply the relevant formulas.

### Step 1: Identify the given data
- Mass of sulfur (solute), \( m_{solute} = 2.56 \, g \)
- Mass of solvent (CS\(_2\)), \( m_{solvent} = 100 \, g \)
- Depression in freezing point, \( \Delta T_f = 0.01 \, ^\circ C \)
- Freezing point depression constant, \( K_f = 0.1 \, ^\circ C \, mol^{-1} \)

### Step 2: Use the freezing point depression formula
The freezing point depression is given by the formula:
\[ \Delta T_f = K_f \cdot m \]
where \( m \) is the molality of the solution.

### Step 3: Calculate molality
Rearranging the formula to find molality:
\[ m = \frac{\Delta T_f}{K_f} \]
Substituting the values:
\[ m = \frac{0.01}{0.1} = 0.1 \, mol/kg \]

### Step 4: Calculate the number of moles of solute
Molality is defined as:
\[ m = \frac{n}{W_{solvent}} \quad \text{(where \( n \) is the number of moles and \( W_{solvent} \) is the weight of solvent in kg)} \]
We need to convert the weight of the solvent from grams to kilograms:
\[ W_{solvent} = 100 \, g = 0.1 \, kg \]
Now we can rearrange the molality formula to find the number of moles \( n \):
\[ n = m \cdot W_{solvent} = 0.1 \cdot 0.1 = 0.01 \, moles \]

### Step 5: Calculate the molar mass of sulfur
The molar mass \( M \) of the solute (sulfur) can be calculated using the formula:
\[ M = \frac{m_{solute}}{n} \]
Substituting the values:
\[ M = \frac{2.56 \, g}{0.01 \, moles} = 256 \, g/mol \]

### Step 6: Determine the atomicity of sulfur
The atomicity of sulfur in CS\(_2\) can be calculated by dividing the molar mass of sulfur by the molar mass of a single sulfur atom:
\[ \text{Atomicity} = \frac{M_{sulfur}}{M_{single \, sulfur}} = \frac{256 \, g/mol}{32 \, g/mol} = 8 \]

### Final Answer
The atomicity of sulfur in CS\(_2\) is **8**.

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