The vapour pressure of `CS_(2)` at `50^(@)C` is `854 "torr"` and a solution of `2.0 g` sulphur in `100 g` of `CS_(2)` has vapour pressure `848.9 "torr"`. If the formula of sulphur molecule is `S_(n)`, then calculate the value of `n`. (at mass of `S = 32`).

Explanation

To solve the problem, we will use Raoult's law, which relates the vapor pressure of a solution to the vapor pressure of the pure solvent and the concentration of the solute.

### Step-by-Step Solution:

1. **Identify Given Data:**
- Vapor pressure of pure CS₂ (P₀) = 854 torr
- Vapor pressure of the solution (Pₛ) = 848.9 torr
- Mass of sulfur (solute) = 2.0 g
- Mass of CS₂ (solvent) = 100 g
- Molar mass of sulfur (S) = 32 g/mol

2. **Calculate the Change in Vapor Pressure:**
\[ \Delta P = P₀ - Pₛ = 854 \, \text{torr} - 848.9 \, \text{torr} = 5.1 \, \text{torr} \]

3. **Use Raoult's Law:**
According to Raoult's law:
\[ \frac{\Delta P}{P₀} = \frac{n_{solute}}{n_{solvent}} \]
where \( n_{solute} \) is the number of moles of the solute and \( n_{solvent} \) is the number of moles of the solvent.

4. **Calculate the Moles of CS₂:**
Molar mass of CS₂ = 12 (C) + 32*2 (S) = 76 g/mol
\[ n_{solvent} = \frac{100 \, \text{g}}{76 \, \text{g/mol}} \approx 1.316 \, \text{mol} \]

5. **Calculate the Moles of Sulfur:**
Let \( M \) be the molar mass of sulfur \( S_n \). The number of moles of sulfur:
\[ n_{solute} = \frac{2.0 \, \text{g}}{M} \]

6. **Substitute Values into Raoult's Law:**
\[ \frac{5.1}{854} = \frac{2.0 / M}{1.316} \]

7. **Cross-Multiply to Solve for M:**
\[ 5.1 \times 1.316 = \frac{2.0}{M} \times 854 \]
\[ 6.71856 = \frac{1708}{M} \]
\[ M = \frac{1708}{6.71856} \approx 254 \, \text{g/mol} \]

8. **Determine the Value of n:**
Since the molar mass of sulfur \( S_n \) is given by:
\[ M = n \times 32 \]
\[ 254 = n \times 32 \]
\[ n = \frac{254}{32} \approx 7.9375 \approx 8 \]

### Final Answer:
The value of \( n \) is approximately **8**.