Question

When 2.0 grams of copper (II) nitrate is added to 1000 ml of pure water, by how much is the vapor pressure of water decreased, given that at 20°C the vapor pressure of pure water is 17.535 mm Hg?

(a) 3.1 x 10^-4

(b) 0.303

(c) 0.0333

(d) 0.0033

This question was addressed to me in my homework.

My doubt stems from Colligative Properties and Determination of Molar Mass in portion Solutions of Chemistry – Class 12

Answer 1

The correct option is (d) 0.0033

Best explanation: Given,

Mass of solute, msolute = 2.0 grams

Volume of solvent, Vsolvent = 1000 ml

P^0water = 17.535 mm Hg

Molecular mass of Cu(NO3)2 = 188 g/mole

We know that ΔP1/P^01 = X2, where X2 is the mole fraction of the solute.

We are required to find out the difference in vapor pressure i.e. ΔP1.

Therefore, ΔPwater= P^0water x X2

Number of moles of copper (II) nitrate = 2.0/188 = 0.0106

Number of moles of water = 1000g/(18g/mole) = 55.56 mole (since density of water is 1g/ml)

Mole fraction of solute, X2 = 0.0106/(0.0106 + 55.56) = 1.9 x 10^-4

ΔPwater = 17.535 x 1.9 x 10^-4 = 0.0033.