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} $)
Show Hint
- 379.2 K
- 377.3 K
- 375.3 K
- 277.3 K
The Correct Option is B
Solution and Explanation
$\Delta T_b = i_1 m_1 k_b + i_2 m_2 k_b$
$\Delta T_b = 1 \times \frac{2}{0.5} \times 0.52 + 1 \times \frac{2}{0.5} \times 0.52 = 4.16$
$(T_b)_{\text{solution}} = 373.16 + 4.16 = 377.3 \text{ K}$
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