5g sulphur is present in 100 g of `CS_(2). DeltaT_(b)` of solution 0.954 and `K_(b)` is 4.88. The molecular formula of sulphur is

Options

A. `S_(2)`
B. `S_(4)`
C. `S_(3)`
D. `S_(8)`

Answer

To solve the problem, we will use the formula for the elevation in boiling point (\( \Delta T_b \)):

\[ \Delta T_b = K_b \cdot m \]

where:
- \( \Delta T_b \) is the elevation in boiling point,
- \( K_b \) is the ebullioscopic constant,
- \( m \) is the molality of the solution.

### Step 1: Calculate the molality (m)

Molality (\( m \)) is defined as:

\[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \]

Given:
- Mass of sulfur (solute) = 5 g
- Mass of \( CS_2 \) (solvent) = 100 g = 0.1 kg

First, we need to find the moles of sulfur:

\[ \text{moles of sulfur} = \frac{\text{mass of sulfur}}{\text{molecular weight of sulfur}} = \frac{5 \text{ g}}{M} \]

where \( M \) is the molecular weight of sulfur.

Now substituting this into the molality formula:

\[ m = \frac{5/M}{0.1} = \frac{50}{M} \]

### Step 2: Substitute into the boiling point elevation formula

Now we can substitute \( m \) into the boiling point elevation formula:

\[ \Delta T_b = K_b \cdot m \]

Substituting the known values:

\[ 0.954 = 4.88 \cdot \frac{50}{M} \]

### Step 3: Solve for M (molecular weight of sulfur)

Rearranging the equation to solve for \( M \):

\[ M = 4.88 \cdot 50 \cdot \frac{1}{0.954} \]

Calculating the right side:

\[ M = \frac{244}{0.954} \approx 255.76 \text{ g/mol} \]

### Step 4: Determine the molecular formula of sulfur

The molecular weight of sulfur is approximately 256 g/mol. The atomic weight of sulfur is approximately 32 g/mol. To find the number of sulfur atoms in the molecular formula:

\[ \text{Number of sulfur atoms} = \frac{256}{32} = 8 \]

Thus, the molecular formula of sulfur is \( S_8 \).

### Final Answer

The molecular formula of sulfur is \( S_8 \).

---