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: Find moles of urea first.
Urea has molar mass $60\ \text{g/mol}$, so moles $= \dfrac{6.0}{60} = 0.1\ \text{mol}$.
Step 2: Convert solvent mass to kg.
Water mass $= 200\ \text{g} = 0.2\ \text{kg}$.
Step 3: Compute molality.
\[ m = \frac{0.1}{0.2} = 0.5\ \text{mol/kg} \]
Step 4: Use boiling-point elevation.
Urea is a non-electrolyte so $i=1$, and \[ \Delta T_b = i\,K_b\,m = 1\times 0.510 \times 0.5 \]
Step 5: Evaluate the rise.
\[ \Delta T_b = 0.255\ \text{K} = 0.255^\circ\text{C} \]
Step 6: Add to the normal boiling point.
\[ T_b = 100 + 0.255 = 100.255^\circ\text{C} \] Splitting moles out first keeps the bookkeeping clean and lands on the same answer. \[ \boxed{T_b = 100.255^\circ\text{C}} \]