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

To determine the boiling point of the solution, we first need to calculate the total elevation in boiling point using the formula for boiling point elevation:

\(\Delta T_b = i \cdot K_b \cdot m\)

where: 
\(\Delta T_b\) is the boiling point elevation, 
\(i\) is the van 't Hoff factor, 
\(K_b\) is the ebullioscopic constant (given as \(0.52 \, \text{K kg mol}^{-1}\)), and 
\(m\) is the molality of the solution.

In this problem, both ethylene glycol (C₂H₆O₂) and glucose (C₆H₁₂O₆) are non-electrolytes, so the van 't Hoff factor \(i = 1\) for both.

The molality \(m\) of the solution is calculated as:

\(m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{2 \, \text{moles} \, (\text{ethylene glycol}) + 2 \, \text{moles} \, (\text{glucose})}{0.5 \, \text{kg}} = \frac{4 \, \text{moles}}{0.5 \, \text{kg}} = 8 \, \text{mol kg}^{-1}\)

Substituting the values into the boiling point elevation formula gives:

\(\Delta T_b = 1 \cdot 0.52 \, \text{K kg mol}^{-1} \cdot 8 \, \text{mol kg}^{-1} = 4.16 \, \text{K}\)

The normal boiling point of water is \(373 \, \text{K}\). Therefore, the boiling point of the solution is:

\(T_b = 373 \, \text{K} + 4.16 \, \text{K} = 377.16 \, \text{K}\)

Rounding this to one decimal place, we get \(377.2 \, \text{K}\), which is closest to the option 377.3 K. Thus, the correct answer is:

377.3 K

Answer 2

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

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

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