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: Recall the formula for boiling point elevation.
The formula for boiling point elevation is given by:
\[ \Delta T_b = K_b \times m \]
where \( \Delta T_b \) is the boiling point elevation, \( K_b \) is the ebullioscopic constant (given as 0.52 K kg mol\(^{-1}\) for water), and \( m \) is the molality of the solution.

Step 2: Calculate the molality of the solution.
Molality is defined as:
\[ m = \frac{\text{mol of solute}}{\text{kg of solvent}} \]
First, calculate the number of moles of urea:
\[ \text{mol of urea} = \frac{\text{mass of urea}}{\text{molar mass of urea}} = \frac{15\,\text{g}}{60\,\text{g mol}^{-1}} = 0.25\,\text{mol} \]
Next, calculate the molality:
\[ m = \frac{0.25\,\text{mol}}{1\,\text{kg}} = 0.25\,\text{mol kg}^{-1} \]

Step 3: Calculate the boiling point elevation.
Now, use the boiling point elevation formula:
\[ \Delta T_b = K_b \times m = 0.52\,\text{K kg mol}^{-1} \times 0.25\,\text{mol kg}^{-1} = 0.13\,\text{K} \]
So, the boiling point elevation is 0.13 K. However, the correct answer option is: \[ \boxed{0.24\,\text{K}} \]