For a dilute solution containing `2.5 g` of a non-volatile non-electrolyte solution in `100g` of water, the elevation in boiling point at `1` atm pressure is `2^(@)C`. Assuming concentration of solute is much lower than the concentration of solvent, the vapour pressure (mm of `Hg)` of the solution is:
(take `k_(b) = 0.76 K kg mol^(-1))`

Options

A. 724
B. 740
C. 736
D. 718

Explanation

To solve the problem of finding the vapor pressure of the solution, we will follow these steps:

### Step 1: Calculate the Molality of the Solution
The elevation in boiling point (ΔTb) is given by the formula:
\[ \Delta T_b = i \cdot k_b \cdot m \]
where:
- \(i\) = van 't Hoff factor (which is 1 for non-electrolytes)
- \(k_b\) = ebullioscopic constant (given as \(0.76 \, K \, kg \, mol^{-1}\))
- \(m\) = molality of the solution (in \(mol/kg\))

Given:
\[ \Delta T_b = 2 \, °C \]
Substituting the known values:
\[ 2 = 1 \cdot 0.76 \cdot m \]
Solving for \(m\):
\[ m = \frac{2}{0.76} \approx 2.63 \, mol/kg \]

### Step 2: Calculate the Moles of Solute
We know that molality \(m\) is defined as:
\[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \]
Given that the mass of the solvent (water) is \(100 \, g\) or \(0.1 \, kg\):
\[ 2.63 = \frac{\text{moles of solute}}{0.1} \]
Thus, the moles of solute:
\[ \text{moles of solute} = 2.63 \cdot 0.1 = 0.263 \, mol \]

### Step 3: Calculate the Molar Mass of the Solute
The mass of the solute is given as \(2.5 \, g\). Therefore, the molar mass \(M\) can be calculated as:
\[ M = \frac{\text{mass of solute}}{\text{moles of solute}} = \frac{2.5 \, g}{0.263 \, mol} \approx 9.50 \, g/mol \]

### Step 4: Calculate the Vapor Pressure of the Solution
Using Raoult's Law, the vapor pressure of the solution can be calculated as:
\[ P_{solution} = P^0_{solvent} - \Delta P \]
Where:
- \(P^0_{solvent}\) is the vapor pressure of pure solvent (water at \(100°C\) is approximately \(760 \, mmHg\)).
- \(\Delta P\) can be calculated using:
\[ \Delta P = \frac{n_{solute}}{n_{solute} + n_{solvent}} \cdot P^0_{solvent} \]
Assuming \(n_{solvent}\) (for \(100 \, g\) of water) is:
\[ n_{solvent} = \frac{100 \, g}{18 \, g/mol} \approx 5.56 \, mol \]
Thus:
\[ \Delta P = \frac{0.263}{0.263 + 5.56} \cdot 760 \approx \frac{0.263}{5.823} \cdot 760 \approx 35.27 \, mmHg \]
Now, substituting back to find \(P_{solution}\):
\[ P_{solution} = 760 - 35.27 \approx 724.73 \, mmHg \]

### Final Answer
The vapor pressure of the solution is approximately \(724.73 \, mmHg\).

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