Question

An aqueous KCl solution of density 1.20 g \(mL^{-1}\) has a molality of 3.30 mol \(kg^{-1}\). The molarity of the solution in mol \(L^{-1}\) is _________. (Nearest integer)
[Molar mass of KCl = 74.5]

Hint:

For quick conversions: \( \text{Molality } (m) = \frac{1000M}{1000d - M \times M_{solute}} \). Memorizing this formula saves time in competitive exams.

Answer 1

Correct Answer: 3

Step 1: Understanding the Concept:
The problem requires converting molality (\(m\)) to molarity (\(M\)) using the density (\(d\)) of the solution and the molar mass of the solute (\(M_{solute}\)). Molality is moles of solute per kilogram of solvent, while molarity is moles of solute per liter of solution.
Step 2: Key Formula or Approach:
The relationship between molarity and molality is given by: \[ M = \frac{1000 \times m \times d}{1000 + (m \times M_{solute})} \]
Alternatively, assume 1 kg of solvent (water). Then: Mass of solvent = 1000 g
Moles of KCl = 3.30 mol
Step 3: Detailed Explanation:
1. Find the total mass of the solution:
Mass of solute (KCl) = \(3.30 \times 74.5 = 245.85 \text{ g}\)
Total mass of solution = \(\text{Mass of solvent} + \text{Mass of solute} = 1000 + 245.85 = 1245.85 \text{ g}\)

2. Find the volume of the solution:
\[ \text{Volume} = \frac{\text{Mass}}{\text{Density}} = \frac{1245.85 \text{ g}}{1.20 \text{ g/mL}} \approx 1038.2 \text{ mL} = 1.0382 \text{ L} \]

3. Calculate Molarity (\(M\)):
\[ M = \frac{\text{moles of solute}}{\text{volume of solution in L}} = \frac{3.30}{1.0382} \approx 3.178 \text{ M} \]
The nearest integer is 3.
Step 4: Final Answer:
The molarity of the solution is 3.