Question

Calculate the boiling point elevation of solution if 15 g urea is dissolved in 1000 g water.
Given:
Given:
\( K_b \, \text{for water} = 0.52 \, \text{K kg mol}^{-1} \), \(\text{molar mass} = 60 \, \text{g mol}^{-1}\), \(\text{mass of urea} = 15 \, \text{g}\), \(\text{mass of water} = 1000 \, \text{g}\)

• A. 0.13 K
• B. 0.24 K
• C. 0.38 K
• D. 0.54 K

Hint:

To calculate boiling point elevation, always use the formula \( \Delta T_b = K_b \times m \), where \( m \) is the molality of the solution.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We need to calculate the elevation in boiling point (\( \Delta\text{T}_\text{b} \)) for a solution containing urea in water.
Step 2: Key Formula or Approach:
Formula: \( \Delta\text{T}_\text{b} = \text{K}_\text{b} \times m \), where \( m \) is molality.
\[ m = \frac{\text{Mass of solute}}{\text{Molar mass of solute} \times \text{Mass of solvent in kg}} \]
Step 3: Detailed Explanation:
Given:
Mass of solute (urea) = 15 g
Molar mass of urea = 60 g/mol
Mass of solvent (water) = 1000 g = 1 kg
\( \text{K}_\text{b} = 0.52 \text{ K kg mol}^{-1} \)
First, calculate molality (\( m \)):
\[ m = \frac{15}{60 \times 1} = \frac{1}{4} = 0.25 \text{ mol/kg} \] Now, calculate boiling point elevation:
\[ \Delta\text{T}_\text{b} = 0.52 \times 0.25 \] \[ \Delta\text{T}_\text{b} = 0.13 \text{ K} \] Step 4: Final Answer:
The boiling point elevation is 0.13 K.