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: Understanding the Question:
We are given the boiling point of a 0.1 molal glucose solution and asked to calculate the boiling point of a 0.5 molal sucrose solution.
Both glucose and sucrose are non-electrolyte solutes, meaning they do not dissociate or associate in an aqueous solution, so the van 't Hoff factor $i = 1$ for both.

Step 2: Key Formula or Approach:
The elevation in boiling point $\Delta T_b$ is a colligative property defined by the formula:
$$\Delta T_b = K_b \times m$$
Where $K_b$ is the molal elevation constant of the solvent (water) and $m$ is the molality of the solution.
The elevated boiling point of the solution is given by $T_b = T_b^\circ + \Delta T_b$, where $T_b^\circ$ is the boiling point of pure water ($100^\circ$C).

Step 3: Detailed Explanation:
Let's first utilize the data for the glucose solution to determine the constant $K_b$.
Given for glucose: $m_1 = 0.1\text{ m}$, $T_{b,1} = 100.16^\circ$C.
The elevation in boiling point for glucose is:
$$\Delta T_{b,1} = T_{b,1} - T_b^\circ = 100.16^\circ\text{C} - 100^\circ\text{C} = 0.16^\circ\text{C}$$
Using the formula:
$$K_b = \frac{\Delta T_{b,1}}{m_1} = \frac{0.16^\circ\text{C}}{0.1\text{ m}} = 1.6^\circ\text{C/m}$$
Now, let's apply this solvent constant to find the elevation in boiling point for the 0.5 molal sucrose solution ($m_2 = 0.5\text{ m}$):
$$\Delta T_{b,2} = K_b \times m_2 = 1.6 \times 0.5 = 0.80^\circ\text{C}$$
Finally, calculate the boiling point of the sucrose solution:
$$T_{b,2} = T_b^\circ + \Delta T_{b,2} = 100^\circ\text{C} + 0.80^\circ\text{C} = 100.80^\circ\text{C}$$

Step 4: Final Answer:
The boiling point of the 0.5 molal sucrose solution is 100.80 $^\circ$C, which perfectly matches option (B).