Question

At $100^{\circ} C$ the vapour pressure of a solution of $6.5 \,g$ of a solute in $100 \,g$ water is $732\, mm$. If $K_b = 0.52$, the boiling point of this solution will be:

• A. $100^{\circ} C$
• B. $102^{\circ} C$
• C. $103^{\circ} C$
• D. $101^{\circ} C$

Answer 1

The Correct Option is D

Boiling Point Elevation Calculation 

The following formula is used to calculate the change in boiling point:

\(\frac{P^{\circ}_A - P_A}{P_A} = \frac{m \times M_A}{1000}\)

Where: - \( P^{\circ}_A \) is the vapor pressure of the pure solvent, - \( P_A \) is the vapor pressure of the solution, - \( m \) is the molality of the solution, - \( M_A \) is the molar mass of the solute (in this case, 18 g/mol for water).

Step 1: Solve for \( m \) (molality)

Substitute the given values into the equation:

\(\frac{760 - 732}{732} = \frac{m \times 18}{1000}\)

Calculating the left-hand side:

\(\frac{760 - 732}{732} = \frac{28}{732} = 0.0382\)

Now solve for \( m \):

\(0.0382 = \frac{m \times 18}{1000}\)

\(m = \frac{0.0382 \times 1000}{18} = 2.125\)

Step 2: Calculate the Boiling Point Elevation

The boiling point elevation \( \Delta T_b \) is calculated using the formula:

\(\Delta T_b = K_b \times m\)

Where \( K_b \) is the ebullioscopic constant of the solvent. Given \( K_b = 0.52 \) for water:

\(\Delta T_b = 0.52 \times 2.125 = 1.10\)

Step 3: Calculate the Final Boiling Point

The final boiling point \( T_s \) is calculated by adding the boiling point elevation to the normal boiling point of the solvent (which is 100°C for water):

\(T_s = 100 + 1.10 = 101.1^{\circ} C\)

Conclusion:

The final boiling point of the solution is \( 101.1^{\circ} C \).