Question

Lowering of vapour pressure due to a solute in 1 molal aqueous solution at ${100}^{o}C$ is:

• A. 13.44 mm Hg
• B. 14.12 mm Hg
• C. 31.2 mm Hg
• D. 35.2 mm Hg

Answer 1

Answer: (A)

13.44 mm Hg

Lowering of vapour pressure = $x_A$

1 molal aqueous solution means 1 mole of solute in 1000 ml of water

density of water = 1 g/ml

in one 1 ml of water = 1 g

in 1000 ml of water = 1000g

moles of water = 55.56 moles

so, $x_A=\frac{n_A}{n_A+n_B}$

so, we get lowering of vapour presssure = 0.01768 atom or 13.43 mm of Hg