For a 1 m glucose aqueous solution (i = 1), boiling point:
A100.52°C (using K_b water = 0.52 K kg/mol)
B100°C
C101°C
D99.48°C
Answer & Solution
Correct answer: A. 100.52°C (using K_b water = 0.52 K kg/mol)
ΔT_b = K_b × m × i = 0.52 × 1 × 1 = 0.52 K. New BP = 100 + 0.52 = 100.52°C.
Keep practising Unit 3B1: Solutions and Concentrations
Browse all Unit 3B1: Solutions and Concentrations questions for JEE Advanced, or explore more Chemistry topics.
Related questions
Hydrogenation of vegetable oils to make margarine uses catalyst:Adsorption isotherm:Freezing-point depression and elevation are colligative properties: depend on:For salt CaCl₂ (assume full dissociation), i =For 0.1 M acetic acid (Ka = 1.8 × 10⁻⁵), van't Hoff factor i is:For solution of urea (M_w = 60 g/mol) with 6 g in 100 g water, molality:Osmotic pressure of 0.01 M sucrose solution at 27°C (R = 0.0821 L atm/mol K):Two volatile liquids A (P°_A = 100) and B (P°_B = 200) form ideal solution. For x_A = 0.4,