Dry air was passed successively through a solution of 5g of a solute in 180g of water and then through pure water and then through pure water. The loss in weight of solution was `2.5g` and that of pure solvent `0.04g`. 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 data provided in the question. Here’s a step-by-step solution:

### Step 1: Understand the problem
We have a solution containing 5g of solute in 180g of water. When dry air is passed through the solution, it causes some of the water to evaporate, leading to a loss in weight of the solution and the pure solvent.

### Step 2: Identify the losses
- Loss in weight of the solution = 2.5g
- Loss in weight of pure solvent (water) = 0.04g

### Step 3: Calculate the loss of water from the solution
The loss of weight of the solution includes the loss of both water and solute. Since we know the loss of pure solvent (water) is 0.04g, we can find the loss of water from the solution:
- Loss of water from the solution = Loss in weight of solution - Loss of weight of pure solvent
- Loss of water from the solution = 2.5g - 0.04g = 2.46g

### Step 4: Calculate the mole fraction of water in the solution
To find the mole fraction of water, we need to calculate the number of moles of water and the number of moles of solute.

1. **Calculate moles of water before evaporation:**
- Molar mass of water (H₂O) = 18 g/mol
- Moles of water = mass of water / molar mass of water
- Moles of water = 180g / 18 g/mol = 10 moles

2. **Calculate moles of water after evaporation:**
- Mass of water after evaporation = 180g - 2.46g = 177.54g
- Moles of water after evaporation = 177.54g / 18 g/mol = 9.86 moles

3. **Calculate moles of solute:**
- Let the molecular weight of the solute be M g/mol.
- Moles of solute = mass of solute / molecular weight of solute = 5g / M

### Step 5: Set up the equation for mole fraction
The mole fraction of water in the solution after evaporation can be expressed as:
\[ \text{Mole fraction of water} = \frac{\text{moles of water after evaporation}}{\text{moles of water after evaporation} + \text{moles of solute}} \]
Substituting the values, we get:
\[ \text{Mole fraction of water} = \frac{9.86}{9.86 + \frac{5}{M}} \]

### Step 6: Use the loss of weight to find the molecular weight
The loss of weight of the solution (2.5g) corresponds to the loss of water and solute. Since we know the loss of water is 2.46g, we can assume that the loss of solute is negligible, and we can focus on the water loss.

Using the relation of vapor pressure lowering (Raoult's Law) and the fact that the loss of weight of the solvent is directly proportional to the mole fraction of the solute, we can set up the equation:
\[ \frac{0.04}{2.46} = \frac{5}{M} \]

### Step 7: Solve for M
Cross-multiplying gives:
\[ 0.04M = 5 \times 2.46 \]
\[ M = \frac{5 \times 2.46}{0.04} = \frac{12.3}{0.04} = 307.5 g/mol \]

### Final Answer
The molecular weight of the solute is approximately **307.5 g/mol**.