Exams › JEE Main › Chemistry

Elevation in the boiling point for 1 molar solution of glucose is 2 K. The depression in the freezing point for 2 molal solution of glucose in the same solvent is 2 K. The relation between Kb and Kf is

1. Kb = Kf
2. Kb = 0.5 Kf
3. Kb = 1.5 Kf
4. Kb = 2 Kf

Correct answer: Kb = 2 Kf

Solution

The relationship between the elevation in boiling point and the depression in freezing point is given by the formulas Kb and Kf, where the change in temperature is directly proportional to the molality of the solution. Since the boiling point elevation for a 1 molal solution of glucose is 2 K and the freezing point depression for a 2 molal solution is also 2 K, it indicates that Kb is twice Kf, leading to the conclusion that Kb = 2 Kf.

Related JEE Main Chemistry questions

• A sulphuric acid solution has a concentration of 3.60 M and contains 29% H2SO4 by mass. If the molar mass of H2SO4 is 98 g mol⁻¹, what is the density of the solution in g mL⁻¹?
• Which one of the following substances is absent in clear hard water?
• Fractional distillation is chosen in a situation where
• Car radiators use an aqueous ethylene glycol solution as antifreeze. How much ethylene glycol should be mixed with 5 kg of water so that the freezing point is lowered to -0.3 °C? (K_f = 1.86 K kg mol⁻¹)
• A solution contains a non-volatile solute of molecular mass M₂. Which expression can be used to determine the molecular mass of the solute using osmotic pressure?
• At 298 K, the Henry’s law constant for oxygen is 1.4 × 10⁻³ mol L⁻¹ atm⁻¹. If oxygen is at a partial pressure of 0.5 atm, what mass of oxygen will dissolve in 100 mL of water at 298 K?

⚔️ Practice JEE Main Chemistry free + battle 1v1 →