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

Step 1: Understanding Molarity and Molality:
Molarity (M): It is defined as the number of moles of solute per liter of solution. \[ M = \frac{\text{moles of solute}}{\text{Volume of solution (L)}} \] Molality (m): It is defined as the number of moles of solute per kilogram of solvent. \[ m = \frac{\text{moles of solute}}{\text{Mass of solvent (kg)}} \] To convert from molarity to molality, we need the density of the solution.
Step 2: Assume a convenient volume and calculate masses:
Let's assume we have 1 Liter of the solution.
Volume of solution = 1 L = 1000 mL.
Mass of the solution: Mass = Volume \(\times\) Density
Mass of solution = 1000 mL \(\times\) 1.86 g/mL = 1860 g.
Moles of the solute: From the given molarity (3 M), 1 L of solution contains 3 moles of solute 'X'.
Moles of solute = 3 mol.
Mass of the solute: Mass = Moles \(\times\) Molar Mass
Mass of solute = 3 mol \(\times\) 120 g/mol = 360 g.
Step 3: Calculate Mass of Solvent and Molality:
Mass of the solvent (water): Mass of solvent = Mass of solution - Mass of solute
Mass of solvent = 1860 g - 360 g = 1500 g.
For molality, we must express the mass of the solvent in kilograms.
Mass of solvent = 1500 g = 1.5 kg.
Calculate molality (m):
\[ m = \frac{\text{moles of solute}}{\text{mass of solvent (in kg)}} = \frac{3 \text{ mol}}{1.5 \text{ kg}} = 2 \text{ mol/kg} \] Thus, the molality of the solution is 2 m.
Step 4: Final Answer:
The molality of the solution is 2m.