Question

What is boiling point of a decimolal aqueous solution of glucose if molal elevation constant for water is $0.52^\circ\text{C kg mol}^{-1}$?

• A. $101.52^\circ\text{C}$
• B. $99.95^\circ\text{C}$
• C. $99.48^\circ\text{C}$
• D. $100.052^\circ\text{C}$

Hint:

Watch out for prefixes! "Molar" (M) refers to moles per Liter of solution, while "Molal" (m) refers to moles per kg of solvent. Colligative property formulas like boiling point elevation strictly use Molality! Decimolal means $0.1\ \text{m}$.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We need to calculate the new boiling point of an aqueous solution of glucose with a specific molality.

Step 2: Key Formula or Approach:
The elevation in boiling point ($\Delta T_b$) is calculated using the formula:
$$\Delta T_b = i \times K_b \times m$$
Where $i$ is the van 't Hoff factor, $K_b$ is the ebullioscopic constant (molal elevation constant), and $m$ is the molality of the solution.
The final boiling point ($T_b$) is the boiling point of the pure solvent plus the elevation:
$$T_b = T_{b}^0 + \Delta T_b$$

Step 3: Detailed Explanation:
The term "decimolal" means a molality of $\frac{1}{10}$, so $m = 0.1\ \text{mol/kg}$.
Glucose is a non-electrolyte and does not dissociate or associate in water, so its van 't Hoff factor is $i = 1$.
The given molal elevation constant for water $K_b = 0.52^\circ\text{C kg mol}^{-1}$.
Now, calculate the boiling point elevation:
$$\Delta T_b = 1 \times 0.52 \times 0.1$$
$$\Delta T_b = 0.052^\circ\text{C}$$
The normal boiling point of pure water ($T_{b}^0$) is exactly $100^\circ\text{C}$.
Calculate the new boiling point:
$$T_b = 100^\circ\text{C} + 0.052^\circ\text{C}$$
$$T_b = 100.052^\circ\text{C}$$

Step 4: Final Answer:
The boiling point of the solution is $100.052^\circ\text{C}$, matching option (D).