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

Answer: (B)

2.79 m

Answer 2

For a 3 M solution, 3 moles of NaCl are present in 1 liter of solution.

The formula for molality $m$ is:

$\text{molality} = \frac{\text{moles of solute} \times 1000}{\text{mass of solvent in grams}}$

Calculate the mass of the solution:

$\text{Mass of solution} = \text{Density} \times \text{Volume} = 1.25 \times 1000 = 1250 \, \text{g}$

Now, calculate the mass of solute (NaCl):

$\text{Mass of solute} = \text{moles} \times \text{molar mass} = 3 \times 58.5 = 175.5 \, \text{g}$

Therefore, the mass of the solvent (water) is:

$\text{Mass of solvent} = 1250 - 175.5 = 1074.5 \, \text{g}$

Substitute the values to find molality:

$\text{molality} = \frac{3 \times 1000}{1074.5} = 2.79 \, m$