Dry air was passed successively through a solution of `5`g of a solute in `180`g water and then through pure water. The loss in weight of solutionwas `250`g and that of pure solvent `0.04`g . The molecular weight of the solute is

Options

A. `31.25`
B. `3.125`
C. `312.5`
D. None of these

Explanation

To find the molecular weight of the solute, we can use the information provided in the problem and apply the concepts of colligative properties, specifically Raoult's law and the concept of vapor pressure lowering.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have a solution containing 5 g of a solute in 180 g of water.
- The loss in weight of the solution when dry air is passed through it is 250 g, and the loss in weight of pure solvent (water) is 0.04 g.

2. **Identify Variables**:
- Let \( W_1 \) = loss of mass of the solution = 250 g
- Let \( W_2 \) = loss of mass of the solvent = 0.04 g
- Let \( W_s \) = mass of solute = 5 g
- Let \( W_w \) = mass of solvent (water) = 180 g

3. **Using the Formula**:
- According to the formula derived from Raoult's law, we have:
\[ \frac{\Delta P}{P} = \frac{W_2}{W_1 + W_2} \]
- Here, \( \Delta P \) is the change in vapor pressure, and \( P \) is the initial vapor pressure.

4. **Substituting the Values**:
- Substitute \( W_1 \) and \( W_2 \) into the equation:
\[ \frac{\Delta P}{P} = \frac{0.04}{250 + 0.04} = \frac{0.04}{250.04} \]

5. **Calculating the Ratio**:
- Calculate the value:
\[ \frac{\Delta P}{P} \approx \frac{0.04}{250} = 0.00016 \]

6. **Using the Formula for Molar Mass**:
- The molar mass of the solute can be calculated using the formula:
\[ \frac{\Delta P}{P} = \frac{W_s}{M \cdot (W_w / M_w)} \]
- Where \( M \) is the molar mass of the solute and \( M_w \) is the molar mass of water (approximately 18 g/mol).

7. **Rearranging the Formula**:
- Rearranging gives us:
\[ M = \frac{W_s \cdot M_w}{W_1 + W_2} \cdot \frac{P}{\Delta P} \]
- Substituting the known values:
\[ M = \frac{5 \cdot 18}{250 + 0.04} \cdot \frac{250.04}{0.04} \]

8. **Calculating Molar Mass**:
- Calculate \( M \):
\[ M = \frac{90}{250.04} \cdot 6251 = 31.25 \text{ g/mol} \]

### Final Answer:
The molecular weight of the solute is approximately **31.25 g/mol**.