Question

Molality $(m)$ of $3 \, \text{M}$ aqueous solution of $\text{NaCl}$ is:
(Given: Density of solution $= 1.25 \, \text{g mL}^{-1}$, Molar mass in $\text{g mol}^{-1}$: $\text{Na} = 23, \, \text{Cl} = 35.5$)

• A. 2.90 m
• B. 2.79 m
• C. 1.90 m
• D. 3.85 m

Answer 1

The Correct Option is B

The molality \(m\) of a 3 M aqueous NaCl solution is calculated using the relationship between molality and molarity. The process is as follows:

1. Known values:
   • Molarity (M) = 3 M
   • Solution density = 1.25 g/mL
   • NaCl molar mass = 23 + 35.5 = 58.5 g/mol
2. Calculate solution mass: A molarity of 3 M indicates 3 moles of NaCl per liter (1000 mL) of solution.
   • Mass of 3 moles of NaCl = \(3 \times 58.5 = 175.5\) g
   • Solution volume = 1000 mL
   • Solution mass = Volume × Density = \(1000 \, \text{mL} \times 1.25 \, \text{g/mL} = 1250 \, \text{g}\)
3. Calculate water mass:
   • Water mass = Solution mass - NaCl mass
   • Water mass = \(1250 - 175.5 = 1074.5 \, \text{g}\)
   • Convert water mass to kg: 1074.5 g = 1.0745 kg
4. Calculate molality: The formula for molality \(m\) is:

\(m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}}\)

• Molality = \(\frac{3 \, \text{moles}}{1.0745 \, \text{kg}} = 2.79 \, \text{m}\)

The molality of the solution is \(2.79\, \text{m}\).