Question

2 moles each of ethylene glycol and glucose are dissolved in 500 g of water. The boiling point of the resulting solution is: (Given: Ebullioscopic constant of water = $ 0.52 \, \text{K kg mol}^{-1} $)

• A. 379.2 K
• B. 377.3 K
• C. 375.3 K
• D. 277.3 K

Hint:

Use the formula for boiling point elevation. Remember to account for the van't Hoff factor (i) for each solute and the total molality of the solution.

Answer 1

The Correct Option is B

$\Delta T_b = i_1 m_1 k_b + i_2 m_2 k_b$

$\Delta T_b = 1 \times \frac{2}{0.5} \times 0.52 + 1 \times \frac{2}{0.5} \times 0.52 = 4.16$

$(T_b)_{\text{solution}} = 373.16 + 4.16 = 377.3 \text{ K}$