Question

The boiling point of 1M aqueous solution of KCl (85% dissociation) having density \(1.04\ \text{g mL}^{-1}\) is \[ \text{(Given: } K_b(\mathrm{H_2O}) = 0.52\ \mathrm{K\ kg\ mol^{-1}}, \quad \text{Molar mass of KCl} = 74.5\ \mathrm{g\ mol^{-1}} \text{)} \]

• A. \(100.096^\circ \mathrm{C}\)
• B. \(100.996^\circ \mathrm{C}\)
• C. \(100.896^\circ \mathrm{C}\)
• D. \(100.796^\circ \mathrm{C}\)

Hint:

For electrolyte colligative-property problems, never use molarity directly in the formula. First convert the given data into molality and then multiply by the appropriate van't Hoff factor. For a salt dissociating into two ions with degree of dissociation \(\alpha\), remember the shortcut: \[ i=1+\alpha \] This saves considerable time in competitive examinations.

Answer 1

The Correct Option is B

Step 1: Understand what is asked.
We have a 1 molar (1M) water solution of KCl. KCl breaks into ions, so it raises the boiling point of water. We want the new boiling point.

Step 2: Pick the right formula.
Boiling point rise is a colligative property, so it depends on the number of particles. The rule is $\Delta T_b = i \, K_b \, m$, where $i$ is the van't Hoff factor, $K_b$ is the molal constant, and $m$ is molality.

Step 3: Find the van't Hoff factor.
One KCl gives 2 ions, so the maximum number of ions per formula is 2. With 85% breaking apart, $i = 1 + \alpha(n-1) = 1 + 0.85(2-1) = 1.85$.

Step 4: Change molarity into molality.
Take 1 litre of solution. Its mass is $1000 \times 1.04 = 1040$ g. The KCl in it weighs $1 \times 74.5 = 74.5$ g. So water mass is $1040 - 74.5 = 965.5$ g $= 0.9655$ kg.
\[ m = \frac{1}{0.9655} = 1.0357 \text{ mol/kg} \]

Step 5: Put values in the formula.
\[ \Delta T_b = 1.85 \times 0.52 \times 1.0357 \]
\[ \Delta T_b = 0.996 \text{ K} \]

Step 6: Add the rise to the normal boiling point.
Pure water boils at $100^\circ$C. So the new boiling point is $100 + 0.996 = 100.996^\circ$C.

Step 7: State the final answer.
The boiling point of the solution is:
\[ \boxed{100.996^\circ \mathrm{C}} \]