Question

The molal boiling point elevation constant for water is \( 0.510 \, \text{K mol}^{-1} \text{ kg} \). The boiling point of a solution made by dissolving \( 6.0 \, \text{g} \) urea in \( 200 \, \text{g} \) water is:

• A. \( 100.255^\circ\text{C} \)
• B. \( 100^\circ\text{C} \)
• C. \( 0.255^\circ\text{C} \)
• D. \( 99.1^\circ\text{C} \)

Hint:

Remember that urea is an organic non-electrolyte, so its van 't Hoff factor \( i \) is always \( 1 \).
Be careful not to select the value of \( \Delta T_b \) (which is \( 0.255^\circ\text{C} \), option C) as the final boiling point. Always add it to \( 100^\circ\text{C} \).

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We are asked to find the boiling point of an aqueous solution containing a known mass of urea dissolved in water, given the molal boiling point elevation constant (\( K_b \)) of water.

Step 2: Key Formula or Approach:
The elevation in boiling point (\( \Delta T_b \)) is given by:
\[ \Delta T_b = i \cdot K_b \cdot m \] where:
- \( i \) is the van 't Hoff factor (for urea, which is a non-electrolyte, \( i = 1 \))
- \( K_b \) is the molal boiling point elevation constant
- \( m \) is the molality of the solution, calculated as:
\[ m = \frac{\text{Moles of solute}}{\text{Mass of solvent in kg}} = \frac{w_2 \times 1000}{M_2 \times w_1} \] where \( w_2 \) is the mass of solute, \( M_2 \) is the molar mass of solute, and \( w_1 \) is the mass of solvent in grams.

Step 3: Detailed Explanation:
Given:
- Mass of solute (urea), \( w_2 = 6.0 \, \text{g} \)
- Molar mass of urea (\( \text{NH}_2\text{CONH}_2 \)), \( M_2 = 60 \, \text{g/mol} \)
- Mass of solvent (water), \( w_1 = 200 \, \text{g} \)
- Molal boiling point elevation constant, \( K_b = 0.510 \, \text{K kg mol}^{-1} \)
First, let's calculate the molality of the solution:
\[ m = \frac{6.0 \times 1000}{60 \times 200} = \frac{6000}{12000} = 0.5 \, \text{mol/kg} \] Now, find the elevation in boiling point:
\[ \Delta T_b = 1 \times 0.510 \times 0.5 = 0.255 \, \text{K} \text{ (or } 0.255^\circ\text{C)} \] Since pure water boils at \( 100^\circ\text{C} \) under normal atmospheric pressure, the boiling point of the solution (\( T_b \)) is:
\[ T_b = T_b^0 + \Delta T_b = 100^\circ\text{C} + 0.255^\circ\text{C} = 100.255^\circ\text{C} \]

Step 4: Final Answer:
(A) \( 100.255^\circ\text{C} \)