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: Write the boiling point rise.
$$\Delta T_b = i\, K_b\, m$$ Glucose does not split, so $i = 1$.

Step 2: Put in the numbers.
Decimolal means $m = 0.1$, and $K_b = 0.52$. So $\Delta T_b = 1 \times 0.52 \times 0.1 = 0.052^\circ\text{C}$.

Step 3: Add to 100.
$$T_b = 100 + 0.052 = 100.052^\circ\text{C}$$
\[ \boxed{100.052^\circ\text{C}} \]