At `100^(@)C` the vapour pressure of a solution of `6.5g` of an solute in `100g` water is `732mm`.If `K_(b)=0.52`, the boiling point of this solution will be :

Options

A. `103^(@)C`
B. `101^(@)C`
C. `100^(@)C`
D. `102^(@)C`

Explanation

To find the boiling point of the solution, we will follow these steps:

### Step 1: Calculate the relative lowering of vapor pressure
The formula for relative lowering of vapor pressure is given by:

\[ \frac{P_0 - P_s}{P_0} \]

Where:
- \( P_0 \) is the vapor pressure of the pure solvent (water at \( 100^\circ C \), which is \( 760 \, \text{mm} \)).
- \( P_s \) is the vapor pressure of the solution, which is given as \( 732 \, \text{mm} \).

Substituting the values:

\[ \frac{760 - 732}{760} = \frac{28}{760} \approx 0.03684 \]

### Step 2: Calculate the number of moles of solute
Using the formula for the number of moles of solute (\( n \)) in terms of the relative lowering of vapor pressure:

\[ \frac{P_0 - P_s}{P_0} = \frac{n}{N} \]

Where \( N \) is the number of moles of solvent. The number of moles of solvent can be calculated using:

\[ N = \frac{\text{mass of solvent (g)}}{\text{molar mass of solvent (g/mol)}} \]

Given:
- Mass of solvent (water) = \( 100 \, \text{g} \)
- Molar mass of water = \( 18 \, \text{g/mol} \)

Calculating \( N \):

\[ N = \frac{100}{18} \approx 5.56 \, \text{moles} \]

Now, substituting back into the equation to find \( n \):

\[ 0.03684 = \frac{n}{5.56} \implies n = 0.03684 \times 5.56 \approx 0.204 \]

### Step 3: Calculate the molality of the solution
Molality (\( m \)) is defined as the number of moles of solute per kilogram of solvent:

\[ m = \frac{n}{\text{mass of solvent (kg)}} \]

Converting the mass of solvent to kg:

\[ \text{mass of solvent} = 100 \, \text{g} = 0.1 \, \text{kg} \]

Now calculating molality:

\[ m = \frac{0.204}{0.1} = 2.04 \, \text{mol/kg} \]

### Step 4: Calculate the boiling point elevation
The boiling point elevation (\( \Delta T_b \)) can be calculated using the formula:

\[ \Delta T_b = K_b \cdot m \]

Where \( K_b \) is the ebullioscopic constant of the solvent (water), given as \( 0.52 \, \text{°C kg/mol} \).

Substituting the values:

\[ \Delta T_b = 0.52 \cdot 2.04 \approx 1.06 \, \text{°C} \]

### Step 5: Calculate the new boiling point of the solution
The new boiling point of the solution is given by:

\[ \text{Boiling point} = 100 + \Delta T_b = 100 + 1.06 = 101.06 \, \text{°C} \]

### Final Answer
The boiling point of the solution is approximately \( 101.06 \, \text{°C} \).

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