Question

A solution is prepared by dissolving 20 g NaOH in 1250 mL of a solvent of density 0.8 g/mL. Then the molality of the solution is

• A. 0.2 mol kg\(^{-1}\)
• B. 0.08 mol kg\(^{-1}\)
• C. 0.25 mol kg\(^{-1}\)
• D. 0.0064 mol kg\(^{-1}\)
• E. 0.5 mol kg\(^{-1}\)

Hint:

Molality uses mass of solvent in kg, not volume of solution.

Answer 1

Answer: (E)

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

Step 1: Calculate moles of NaOH.
Molar mass of NaOH: \[ 40\ \text{g mol}^{-1} \] \[ \text{Moles}=\frac{20}{40}=0.5 \]

Step 2: Calculate mass of solvent.
\[ \text{Mass}=\text{volume}\times\text{density} \] \[ =1250\times0.8=1000\ \text{g}=1\ \text{kg} \]

Step 3: Calculate molality.
\[ m=\frac{0.5}{1}=0.5\ \text{mol kg}^{-1} \]