Question

Calculate the molality of 3.0 g acetic acid in 100 g of benzene.

• A. $0.05 \ \text{mol kg}^{-1}$
• B. $0.5 \ \text{mol kg}^{-1}$
• C. $2.0 \ \text{mol kg}^{-1}$
• D. $0.2 \ \text{mol kg}^{-1}$

Hint:

Molality depends only on mass (not temperature), so it is preferred in colligative property calculations.

Answer 1

The Correct Option is B

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

Step 1: Find moles of acetic acid.
Molar mass of acetic acid ($CH_3COOH$) = $60 \ \text{g/mol}$ \[ \text{Moles} = \frac{3.0}{60} = 0.05 \ \text{mol} \]

Step 2: Convert mass of solvent into kg.
\[ 100 \ \text{g} = 0.1 \ \text{kg} \]

Step 3: Calculate molality.
\[ m = \frac{0.05}{0.1} = 0.5 \ \text{mol kg}^{-1} \]

Step 4: Final conclusion.
Thus, the molality is: \[ 0.5 \ \text{mol kg}^{-1} \]