Question

how many grams of urea should add to 100 grams of pure water to minimize the vapour pressure of water to 0.5 kpa in 0 centigrade of tempreture.

Answer 1

Experimentally, we know that the vapor pressure of the solvent above a solution containing a non-volatile solute (i.e., a solute that does not have a vapor pressure of its own) is directly proportional to the mole fraction of solvent in the solution. This behavior is summed up in Raoult's Law:

$$P_{\text{solvent}} = X_{\text{solvent}} P_{\text{solvent}}^{\circ}$$

where:

$P_{\text{solvent}}$ is the vapor pressure of the solvent above the solution,

$X_{\text{solvent}}$ is the mole fraction of the solvent in the solution,

$P_{\text{solvent}}^{\circ}$ is the vapor pressure of the pure solvent.

From reference data $P^0 = 0.6\ \mathrm{kPa}$ at $0\ \mathrm{C}$

P must be $0.5\ \mathrm{kPa}$, so X is next:

$$0.5 = x \times 0.6$$ $$x = 0.8333, \text{ so}$$

n of water $= 100/18 = 5.55\ \mathrm{mol}$

n of urea $= x/(5.55+x) = (1 - 0.8333)$

$$x = 1,11\ \mathrm{mol}$$

Mw of urea $= 60$

$$m = n \times Mw = 60 \times 1,11 = 6.66\ \mathrm{g}$$