The properties of solutions which depend on the number of solute particles and are independent of their chemical identity are called colligative properties. These are lowering of vapour pressure, elevation of boiling point, depression of freezing point and osmotic pressure. The process of osmosis can be reversed if a pressure higher than the osmotic pressure is applied to the solution. Colligative properties have been used to determine the molar mass of solutes. Solutes which dissociate in solution exhibit molar mass lower than the actual molar mass and those which associate show higher molar mass than their actual values. Quantitatively, the extent to which a solute is dissociated or associated can be expressed by van’t Hoff factor i. This factor has been defined as ratio of normal molar mass to experimentally determined molar mass or as the ratio of observed colligative property to the calculated colligative property.

18 g of glucose, C₆H₁₂O₆ (Molar mass: 180 g mol^-1) is dissolved in 1 kg of water in a saucepan. At what temperature will this solution boil? (K_b for water = 0.52 K kg mol^-1, boiling point of pure water = 373.15 K)

373.202 K

The correct answer is option 3. 373.202K.

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

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

where:

\(\Delta T_b\) is the boiling point elevation,

\(i\) is the van't Hoff factor (for glucose, which does not dissociate in water, \(i = 1\)),

\(K_b\) is the ebullioscopic constant (boiling point elevation constant) for the solvent (water in this case),

\(m\) is the molality of the solution.

First, we calculate the molality (\(m\)) of the solution:

\(m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}}\)

The moles of glucose can be found by dividing the mass of glucose by its molar mass:

\(\text{moles of glucose} = \frac{18 \text{ g}}{180 \text{ g mol}^{-1}} = 0.1 \text{ mol}\)

The mass of the solvent (water) is 1 kg. So, the molality (\(m\)) is:

\(m = \frac{0.1 \text{ mol}}{1 \text{ kg}} = 0.1 \text{ mol kg}^{-1}\)

Next, we calculate the boiling point elevation (\(\Delta T_b\)):

\(\Delta T_b = 1 \times 0.52 \text{ K kg mol}^{-1} \times 0.1 \text{ mol kg}^{-1} = 0.052 \text{ K}\)

The boiling point of the solution is the boiling point of pure water plus the boiling point elevation:

\(T_{\text{solution}} = 373.15 \text{ K} + 0.052 \text{ K} = 373.202 \text{ K}\)

Thus, the temperature at which this solution will boil is: 373.202 K