The molality of a solution prepared by dissolving 4.0 g NaOH in 1000 mL distilled water is _____

(Assume that the density of water is $1\, g\, mL^{-1}$)

$0.1\, mol\, kg^{-1}$

The correct answer is Option (4) → $0.1\, mol\, kg^{-1}$

We are asked to calculate the molality (m):

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

Step 1: Calculate moles of NaOH

$\text{Molar mass of NaOH} = 23 + 16 + 1 = 40 \, \text{g/mol}$

$\text{Moles of NaOH} = \frac{4.0 \, \text{g}}{40 \, \text{g/mol}} = 0.1 \, \text{mol}$

Step 2: Mass of solvent in kg

• Volume of water = 1000 mL = 1 L
• Density = 1 g/mL
• Mass of water = 1000 × 1 = 1000 g=1kg

Step 3: Molality

m = moles of solute / mass of solvent (kg)

$m = \frac{0.1 \, \text{mol}}{1 \, \text{kg}} = 0.1 \, \text{mol/kg}$