The vapour pressure of solution containing 5 g of non-electrolyte in 90g of water at a particular temperature is 2985 `Nm^(–2)`. The vapour pressure of pure water at that temperature is 3000 `Nm^(–2)`. The molecular weight of the solute is –

Options

A. 180
B. 90
C. 270
D. 200

Answer

To find the molecular weight of the solute in the given solution, we can follow these steps:

### Step 1: Identify the given data
- Mass of solute (non-electrolyte) = 5 g
- Mass of solvent (water) = 90 g
- Vapor pressure of the solution (P_solution) = 2985 Nm⁻²
- Vapor pressure of pure water (P_pure) = 3000 Nm⁻²

### Step 2: Calculate the relative lowering of vapor pressure
The relative lowering of vapor pressure (RLVP) can be calculated using the formula:
\[ \text{RLVP} = \frac{P_{\text{pure}} - P_{\text{solution}}}{P_{\text{pure}}} \]
Substituting the values:
\[ \text{RLVP} = \frac{3000 - 2985}{3000} = \frac{15}{3000} = 0.005 \]

### Step 3: Relate RLVP to the mole fraction of the solute
According to Raoult's law, the relative lowering of vapor pressure is equal to the mole fraction of the solute (x_B):
\[ \text{RLVP} = x_B \]
Thus, we have:
\[ x_B = 0.005 \]

### Step 4: Write the expression for mole fraction
The mole fraction of the solute (x_B) can be expressed as:
\[ x_B = \frac{n_B}{n_A + n_B} \]
Where:
- \(n_B\) = number of moles of solute
- \(n_A\) = number of moles of solvent (water)

Since the number of moles of solute is much smaller than that of the solvent, we can approximate:
\[ x_B \approx \frac{n_B}{n_A} \]

### Step 5: Calculate the number of moles of solvent
The number of moles of water (n_A) can be calculated using:
\[ n_A = \frac{\text{mass of water}}{\text{molar mass of water}} = \frac{90 \text{ g}}{18 \text{ g/mol}} = 5 \text{ mol} \]

### Step 6: Substitute to find the number of moles of solute
Using the mole fraction:
\[ 0.005 = \frac{n_B}{5} \]
Thus,
\[ n_B = 0.005 \times 5 = 0.025 \text{ mol} \]

### Step 7: Calculate the molecular weight of the solute
The molecular weight (M) of the solute can be calculated using:
\[ M = \frac{\text{mass of solute}}{n_B} = \frac{5 \text{ g}}{0.025 \text{ mol}} = 200 \text{ g/mol} \]

### Final Answer
The molecular weight of the solute is **200 g/mol**.

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