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

Concept: The elevation in boiling point is a colligative property. Colligative properties depend only on the number of solute particles present in the solution and not on their chemical nature. When an electrolyte such as KCl dissolves in water, it dissociates into ions and hence produces more particles than a non-electrolyte. For boiling point elevation, the relation is \[ \Delta T_b=iK_bm \] where

• \( \Delta T_b \) = elevation in boiling point,

• \( i \) = van't Hoff factor,

• \( K_b \) = ebullioscopic constant,

• \( m \) = molality of the solution.
Since KCl dissociates according to \[ \mathrm{KCl \rightarrow K^+ + Cl^-} \] the maximum number of particles formed from one formula unit is 2. However, the dissociation is only \(85\%\), therefore the actual van't Hoff factor must be calculated using the degree of dissociation. To determine the boiling point accurately, we first calculate the mass of solvent present in one litre of solution and then determine the molality.

Step 1: Calculating the van't Hoff factor \(i\). For an electrolyte producing two ions, the van't Hoff factor is \[ i=1+\alpha(n-1) \] where \[ \alpha=0.85 \] and \[ n=2 \] Substituting the values, \[ i=1+0.85(2-1) \] \[ i=1+0.85 \] \[ i=1.85 \] Thus, due to \(85\%\) dissociation, every mole of KCl effectively behaves as \(1.85\) moles of particles.

Step 2: Determining the mass of one litre of solution. The molarity is given as \(1\,M\). Therefore, \[ 1\ \text{litre of solution contains }1\ \text{mole of KCl}. \] Density of solution is \[ 1.04\ \text{g mL}^{-1} \] Hence mass of \(1000\) mL solution is \[ 1000 \times 1.04 \] \[ =1040\ \text{g} \] Therefore, \[ \text{Mass of solution}=1040\ \text{g} \]

Step 3: Calculating the mass of KCl present. Since the solution is \(1M\), \[ 1\ \text{mole of KCl is present in }1\ \text{L solution} \] Given molar mass of KCl \[ =74.5\ \text{g mol}^{-1} \] Therefore, \[ \text{Mass of KCl}=74.5\ \text{g} \]

Step 4: Calculating the mass of solvent (water). Mass of solvent \[ =\text{Mass of solution} - \text{Mass of solute} \] \[ =1040-74.5 \] \[ =965.5\ \text{g} \] Converting into kilograms, \[ 965.5\ \text{g} = 0.9655\ \text{kg} \] Thus, \[ \text{Mass of water}=0.9655\ \text{kg} \]

Step 5: Calculating molality of the solution. Molality is defined as \[ m=\frac{\text{moles of solute}}{\text{kg of solvent}} \] Substituting the values, \[ m=\frac{1}{0.9655} \] \[ m\approx1.036 \] Hence, \[ m=1.036\ \text{mol kg}^{-1} \]

Step 6: Calculating elevation in boiling point. Using \[ \Delta T_b=iK_bm \] Substituting all known values, \[ \Delta T_b = (1.85)(0.52)(1.036) \] First calculate \[ 1.85\times0.52 = 0.962 \] Then, \[ 0.962\times1.036 = 0.996 \] Therefore, \[ \Delta T_b = 0.996\ \mathrm{K} \]

Step 7: Determining the boiling point of the solution. The normal boiling point of pure water is \[ 100^\circ \mathrm{C} \] Therefore, \[ \text{Boiling point} = 100 + \Delta T_b \] \[ = 100+0.996 \] \[ = 100.996^\circ \mathrm{C} \]

Final Answer: \[ \boxed{100.996^\circ \mathrm{C}} \] Hence, the correct option is \[ \boxed{\text{(B) }100.996^\circ \mathrm{C}} \]