Question

The density of \( 3\,M \) aqueous solution of a solute \( X \) is \( 1.86\,\text{g mL}^{-1} \). The molality of the solution is (Molar mass of solute \( X \) is \( 120\,\text{g mol}^{-1} \))

• A. \(3\,m \)
• B. \(4\,m \)
• C. \(2\,m \)
• D. \(5\,m \)
• E. \(1\,m \)

Hint:

Use 1 L solution to convert molarity → molality easily.

Answer 1

The Correct Option is C

Concept: \[ \text{Molality} = \frac{\text{moles of solute}}{\text{kg of solvent}} \]

Step 1: Take 1 L solution.
Moles of solute = 3 mol Mass of solution: \[ = 1.86 \times 1000 = 1860\,\text{g} \] Mass of solute: \[ = 3 \times 120 = 360\,\text{g} \]

Step 2: Mass of solvent.
\[ = 1860 - 360 = 1500\,\text{g} = 1.5\,\text{kg} \]

Step 3: Molality.
\[ m = \frac{3}{1.5} = 2\,m \]