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. (a) `31.25`
B. (b) `3.125`
C. (c ) `312.5`
D. (d) None of these

Explanation

To find the molecular weight of the solute, we can use the information provided in the question and apply the concept of colligative properties, specifically Raoult's law. Here’s a step-by-step solution:

### Step 1: Understand 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

### Step 2: Apply Raoult's Law
According to Raoult's law, the relative lowering of vapor pressure is proportional to the mole fraction of the solute in the solution. This can be expressed as:

\[ \frac{P_0 - P_s}{P_0} = \text{mole fraction of solute} \]

Where:
- \(P_0\) = vapor pressure of pure solvent
- \(P_s\) = vapor pressure of the solution

The change in vapor pressure can be related to the loss in weight of the solvent and the solution:

\[ \frac{\Delta W_{solvent}}{\Delta W_{solution}} = \frac{P_0 - P_s}{P_0} \]

### Step 3: Set Up the Ratio
Using the given loss in weight:

\[ \frac{0.04}{2.50} = \frac{n_{solute}}{n_{solvent}} \]

Where:
- \(n_{solute}\) = number of moles of solute
- \(n_{solvent}\) = number of moles of solvent

### Step 4: Calculate Moles of Solvent
The number of moles of solvent (water) can be calculated using its mass and molar mass:

\[ n_{solvent} = \frac{W_{solvent}}{M_{solvent}} = \frac{180 g}{18 g/mol} = 10 mol \]

### Step 5: Express Moles of Solute
Let \(M_{solute}\) be the molar mass of the solute. The number of moles of solute can be expressed as:

\[ n_{solute} = \frac{W_{solute}}{M_{solute}} = \frac{5 g}{M_{solute}} \]

### Step 6: Substitute and Rearrange
Substituting the expressions for moles into the ratio:

\[ \frac{0.04}{2.50} = \frac{5/M_{solute}}{10} \]

This simplifies to:

\[ \frac{0.04}{2.50} = \frac{5}{10 M_{solute}} \]

### Step 7: Solve for Molar Mass
Cross-multiplying gives:

\[ 0.04 \times 10 M_{solute} = 5 \times 2.50 \]

Calculating the right side:

\[ 0.4 M_{solute} = 12.5 \]

Now, solve for \(M_{solute}\):

\[ M_{solute} = \frac{12.5}{0.4} = 31.25 g/mol \]

### Conclusion
The molecular weight of the solute is **31.25 g/mol**.

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