Question

At 70°C the vapor pressure of pure water is 31 kPa. Which of the following is most likely the vapor pressure of a 2.0 molal aq. glucose solution at 70°C?

(a) 30.001 kPa

(b) 29.915 kPa

(c) 28.226 kPa

(d) 32.392 kPa

This question was addressed to me in class test.

Query is from Colligative Properties and Determination of Molar Mass topic in portion Solutions of Chemistry – Class 12

Answer 1

The correct option is (b) 29.915 kPa

The best explanation: Given, P^0water = 31 kPa

Concentration of solution, c = 2 molal = 2 moles of glucose/kg of water

From law of relative lowering of vapor pressure, ΔP/P^0 = X2, where X2 is the mole fraction of glucose in the solution.

Mass of water = 1 kg = 1000 g

Molecular weight of water = 18 g/mole

Moles of water = 1000/18 = 55.556 moles

X2 = 2/(2 + 55.556) = 0.035

ΔP = 31 kPa x 0.035 = 1.085 kPa

Final pressure = 31 kPa – 1.085 kPa = 29.915 kPa.