Vapour pressure of a solution of `5g`of non-electrolyte in `100g`water at a particular temperature is`2985N//m^(2)`.The vapour pressure of pure water is`3000N//m^(2)`.The molecular weight of the solute is

Options

A. 60. 0
B. 120. 0
C. 180. 0
D. 380. 0

Explanation

To find the molecular weight of the non-electrolyte solute, we can follow these steps:

### Step 1: Understand the Given Data
- Mass of solute (non-electrolyte) = 5 g
- Mass of solvent (water) = 100 g
- Vapour pressure of the solution (Pt) = 2985 N/m²
- Vapour pressure of pure water (P0) = 3000 N/m²

### Step 2: Calculate the Mole Fraction of the Solvent
The formula for the vapour pressure of a solution is given by Raoult's Law:
\[ P_t = P_0 \cdot X_{solvent} \]
Where:
- \( P_t \) = Vapour pressure of the solution
- \( P_0 \) = Vapour pressure of the pure solvent
- \( X_{solvent} \) = Mole fraction of the solvent

Rearranging the formula to find the mole fraction of the solvent:
\[ X_{solvent} = \frac{P_t}{P_0} = \frac{2985}{3000} \]

Calculating:
\[ X_{solvent} = 0.995 \]

### Step 3: Calculate the Mole Fraction of the Solute
Since the sum of the mole fractions of the solute and solvent equals 1:
\[ X_{solute} = 1 - X_{solvent} = 1 - 0.995 = 0.005 \]

### Step 4: Calculate the Moles of Solvent
The number of moles of water can be calculated using its molar mass (approximately 18 g/mol):
\[ \text{Moles of water} = \frac{\text{mass of water}}{\text{molar mass of water}} = \frac{100 \, g}{18 \, g/mol} \approx 5.56 \, mol \]

### Step 5: Set Up the Mole Fraction Equation
The mole fraction of the solute is given by:
\[ X_{solute} = \frac{\text{moles of solute}}{\text{moles of solute} + \text{moles of solvent}} \]
Let \( m \) be the molecular weight of the solute. The moles of solute can be expressed as:
\[ \text{Moles of solute} = \frac{5 \, g}{m \, g/mol} \]

Substituting into the mole fraction equation:
\[ 0.005 = \frac{\frac{5}{m}}{\frac{5}{m} + 5.56} \]

### Step 6: Solve for m
Cross-multiplying gives:
\[ 0.005 \left(\frac{5}{m} + 5.56\right) = \frac{5}{m} \]

Expanding and rearranging:
\[ 0.005 \cdot 5.56 = \frac{5}{m} - 0.005 \cdot \frac{5}{m} \]
\[ 0.0278 = \frac{5}{m} (1 - 0.005) \]
\[ 0.0278 = \frac{5}{m} \cdot 0.995 \]
\[ \frac{5}{m} = \frac{0.0278}{0.995} \]

Calculating:
\[ \frac{5}{m} \approx 0.0279 \]
\[ m \approx \frac{5}{0.0279} \approx 179.57 \, g/mol \]

### Step 7: Conclusion
The molecular weight of the solute is approximately **180 g/mol**.