2.56 g of suphur in 100 g of `CS_(2)` has depression in freezing point of `0.010^(@)C , K_(f)=0.1^(@)C` `(molal)^(-1)`. Hence atomicity of sulphur in the solution is

Options

A. 2
B. 4
C. 6
D. 8

Answer

To solve the problem, we need to determine the atomicity of sulfur in the solution based on the given data. Here’s the step-by-step solution:

### Step 1: Understand the Given Data
We have:
- Mass of sulfur (solute) = 2.56 g
- Mass of CS₂ (solvent) = 100 g
- Depression in freezing point (ΔTf) = 0.010 °C
- Freezing point depression constant (Kf) = 0.1 °C/molal

### Step 2: Define Atomicity of Sulfur
Let the atomicity of sulfur be \( N \). Therefore, sulfur can be represented as \( S_N \), where \( N \) is the number of sulfur atoms in the molecule.

### Step 3: Calculate Molar Mass of Sulfur
The molar mass of the solute \( S_N \) can be expressed as:
\[ \text{Molar mass of } S_N = 32N \text{ g/mol} \]
where the atomic mass of sulfur is 32 g/mol.

### Step 4: Use the Freezing Point Depression Formula
The formula for depression in freezing point is given by:
\[ \Delta T_f = K_f \times m \]
where \( m \) is the molality of the solution.

### Step 5: Calculate Molality
Molality \( m \) is defined as:
\[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \]
The number of moles of solute can be calculated as:
\[ \text{moles of solute} = \frac{\text{mass of solute}}{\text{molar mass of solute}} = \frac{2.56}{32N} \]
The mass of the solvent in kg is:
\[ \text{mass of solvent} = 100 \text{ g} = 0.1 \text{ kg} \]
Thus, the molality can be expressed as:
\[ m = \frac{2.56 / (32N)}{0.1} = \frac{25.6}{32N} \]

### Step 6: Substitute into the Freezing Point Depression Formula
Substituting the expression for molality into the freezing point depression formula:
\[ 0.010 = 0.1 \times \frac{25.6}{32N} \]

### Step 7: Solve for \( N \)
Rearranging the equation gives:
\[ 0.010 = \frac{0.1 \times 25.6}{32N} \]
Multiplying both sides by \( 32N \):
\[ 0.010 \times 32N = 0.1 \times 25.6 \]
\[ 0.32N = 2.56 \]
Now, solving for \( N \):
\[ N = \frac{2.56}{0.32} = 8 \]

### Conclusion
The atomicity of sulfur in the solution is \( N = 8 \).

### Final Answer
The atomicity of sulfur in the solution is **8**.

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