Question

What is the boiling point of 0.5 molal aqueous solution of sucrose if 0.1 molal aqueous solution of glucose boils at 100.16 $^\circ$C?

• A. 100.32 $^\circ$C
• B. 100.80 $^\circ$C
• C. 100.16 $^\circ$C
• D. 100.62 $^\circ$C

Hint:

Since both solutes are non-electrolytes in the same solvent, the elevation in boiling point is directly proportional to the molality ($\Delta T_b \propto m$).
If the molality increases by 5 times (from 0.1 m to 0.5 m), the boiling point elevation must also increase by 5 times: $0.16^\circ\text{C} \times 5 = 0.80^\circ\text{C}$.
Add this directly to $100^\circ$C to get $100.80^\circ$C in seconds!

Answer 1

The Correct Option is B

Step 1: Understand the question.
A 0.1 molal glucose solution boils at 100.16 degrees. We must find the boiling point of a 0.5 molal sucrose solution. Both solutes do not split up, so each behaves simply.
Step 2: Recall the boiling point rule.
The rise in boiling point is: \[ \Delta T_b = K_b \times m \] where $K_b$ is a constant for water and $m$ is the molality.
Step 3: Find the rise for glucose.
\[ \Delta T_b = 100.16 - 100 = 0.16 \text{ degrees} \]
Step 4: Find the constant.
\[ K_b = \frac{\Delta T_b}{m} = \frac{0.16}{0.1} = 1.6 \]
Step 5: Find the rise for sucrose.
\[ \Delta T_b = 1.6 \times 0.5 = 0.80 \text{ degrees} \] So the boiling point is $100 + 0.80 = 100.80$ degrees.
Step 6: Choose the answer.
The boiling point is 100.80 degrees, which is option 2. \[ \boxed{100.80\,^\circ\text{C}} \]