This question concerns freezing point depression and boiling point elevation.

Question

The specific gravity of a substance is defined as the ratio of the substance's density to the density of water. For example, a substance with a specific gravity of 0.750 is 75.0% as dense as water, and thus has a density of 0.750 x 0.998 g/mL=0.749 g/mL. A solution is prepared by adding carbon tetrachloride, CCl4, to 2.50 L of benzene. The specific gravity of benzene is 0.878. If the boiling point of the solution is found to be 85.34 degrees C, determine the mass in grams of the CCl4 in the solution. 

Answer 1

Come on, Ivy E.,

Have you posted your entire physical chemistry problem set to Quora (4 problems, same general area, at the same time)? That's not the intent of Quora, you know. If you are "at sea" with respect to the whole area of colligative properties, you would do better to get one-on-one tutoring, so that you can learn the reasoning required, rather than try to absorb posted solutions....

-- Cheers, --Mr. d.

Answer 2

Before approaching this problem, we need to know the normal boiling point for benzene as well as the Kb (molal boiling point elevation constant) for benzene. The values used below may vary from the values you have. Adjust calculations accordingly.

Normal bp of benzene = 80.08ºC (tabular value from internet)

Kb (molal boiling point elevation constant = 2.64º/m (tabular value from internet)

∆T = imK

∆T = change in boiling point = 85.34º - 80.08º = 5.26º

i = van't Hoff factor for CCl₄ = 1 (non electrolyte)

m = molality = moles CCl₄ / kg benzene = ?

K = boiling constant for benzene = 2.64º/m

Solving for m (molality) which is moles of CCl₄ / kg of benzene, we have...

5.26 = (1)(m)(2.64)

m = 1.99 mols CCl₄ / kg benzene

Since we have 2.50 L of benzene, we can determine kg of benzene using specific gravity of benzene:

density of benzene = 0.878 x 0.998 g / ml = 0.876 g / ml

2.50 L x 1000 ml / L = 2500 ml x 0.876 g / ml = 2190 g of benzene = 2.191 kg of benzene

1.99 mols CCl₄ / kg benzene x 2.190 kg benzene = 4.359 mols CCl₄

Molar mass CCl₄ = 153.8 g / mol

Mass of CCl₄ = 4.359 mols x 153.8 g / mol = 670. g CCl₄