Dry air was passed successively through solution of `5g` of a solute in `180g` of water and then through pure water. The loss in weight of solution was `2.50 g` and that of pure solvent `0.04g`. The molecualr weight of the solute is:

Options

A. 31.25
B. 3.125
C. 312.5
D. None

Explanation

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

### Step-by-Step Solution

1. **Identify the Given Data:**
- Mass of solute (W_solute) = 5 g
- Mass of solvent (W_solvent) = 180 g
- Loss in weight of solution (ΔW_solution) = 2.50 g
- Loss in weight of pure solvent (ΔW_solvent) = 0.04 g

2. **Calculate the Total Loss in Weight:**
The total loss in weight when dry air passes through both the solution and the pure solvent is:
\[ \Delta W_{total} = \Delta W_{solution} + \Delta W_{solvent} = 2.50 \, \text{g} + 0.04 \, \text{g} = 2.54 \, \text{g} \]

3. **Calculate the Relative Lowering of Vapor Pressure:**
The relative lowering of vapor pressure (R) can be calculated using the formula:
\[ R = \frac{\Delta W_{solvent}}{\Delta W_{total}} = \frac{0.04}{2.54} \]

4. **Calculate the Moles of Solute:**
Let the molecular weight of the solute be \( M \) g/mol. The number of moles of solute (n_solute) is given by:
\[ n_{solute} = \frac{W_{solute}}{M} = \frac{5}{M} \]

5. **Calculate the Moles of Solvent:**
The molecular weight of water is approximately 18 g/mol. Therefore, the number of moles of solvent (n_solvent) is:
\[ n_{solvent} = \frac{W_{solvent}}{18} = \frac{180}{18} = 10 \, \text{moles} \]

6. **Calculate the Mole Fraction of Solute:**
The mole fraction of the solute (X_solute) is given by:
\[ X_{solute} = \frac{n_{solute}}{n_{solute} + n_{solvent}} = \frac{\frac{5}{M}}{\frac{5}{M} + 10} \]

7. **Set Up the Equation:**
From the relative lowering of vapor pressure, we know:
\[ R = X_{solute} = \frac{0.04}{2.54} \]
Therefore, we can set up the equation:
\[ \frac{\frac{5}{M}}{\frac{5}{M} + 10} = \frac{0.04}{2.54} \]

8. **Cross-Multiply and Solve for M:**
Cross-multiplying gives:
\[ 5 \cdot 2.54 = 0.04 \cdot (5 + 10M) \]
Simplifying this:
\[ 12.7 = 0.04 \cdot 5 + 0.04 \cdot 10M \]
\[ 12.7 = 0.2 + 0.4M \]
\[ 12.5 = 0.4M \]
\[ M = \frac{12.5}{0.4} = 31.25 \, \text{g/mol} \]

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