A.) Water at 0°C was placed in a dish inside a vessel maintained at low pressure

Question

A.)

Water at 0°C was placed in a dish inside a vessel maintained at low pressure by a vacuum pump. After a quantity of water has evaporated, the remainder froze. If 6.30 g of ice at 0° C was obtained, how many grams of liquid water must have evaporated? The heat of fusion of water is 6.01 kJ/mol and the heat of vaporization is 44.9 kJ/mol.

B.)

Would you expect CCl4 to be soluble in H2O? (Hint: Begin your explanation by identifying the intermolecular forces of attraction of the solute and solvent.)

C.)

A 2.15 g sample of molecular solid was dissolved in 25.0 g of CS2. The boiling point elevation of the solution was 1.59°C. Find the molar mass of the unknown. (Kb of CS2 is 1.07°C/m)

D.)

The normal melting and boiling points of krypton are -157°C and -152°C, respectively, and its triple point is at -169°C and .175 atm. Sketch the phase diagram for krypton, showing the three points given above and indicating the areas in which each phase is stable.

Explanation / Answer

Answer:

A).

Energy lost by the freezing 6.30grams of water is equal to the energy gained by vaporizing ’ X ‘ grams of water.

Molecular weight of water = 18 grams/mole

Given weight of water = 6.30 grams

Latent heat of vaporization = 44.9 kJ/Mole
(6.30 * 6.01 * 18) = X * (44.9 * 18)
(6.30) * 6.01 = X * 44.9
Weight = (6.30*6.01)/ 44.9 = 0.845 grams

B). CCl4 is non-polar molecule in nature and H2O is polar molecule in nature. The extent of solubility of CCl4 in H2O is very less.

C).

Tb = 1.59 0C

Kb = 1.07 0C/m

Given weight of sample = 2.15 grams

Given weight of solvent = 25 grams

1.59 = (1.07) *[(2.15)/(Molecular weight)/(25*10-3kg)] * (1)
Molecular weight = (1.07) *[(2.15)/(1.59)/(25*10-3kg)]
Molecular weight of unknown sample = 57.87 grams

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