A student determines the molar mass of acetone, CH3COCH3, by the method used in this experiment. She found that the equilibrium temperature of a mixture of ice and water was 1.0 degrees C on her thermometer. When she added 11.1 g of her sample to the mixture, the temperature, after thorough stirring, fell to -3.0 degrees C. She then poured off the solution through a screen into a beaker. The mass of the solution was 90.4 g. a) What was the freezing point depression?b) What was the molality of acetone? I need help especially with this one. c) How much acetone was in the decanted solution?d) How much water is in the decanted solution?e) How much acetone would there be in a solution containing 1 kg of water and acetone at the same concentration as she had in her experiment?f) What did she find to be the molar mass of acetone, assuming she made the calculation properly?

(a). 4°C, (b). 2.4M, (c). 11.1 g, (d). 89.01 g, (e). 139.2 g and (f). 58 g/mol.

Explanation:

Without mincing words let's dive straight into the solution to the question.

(a). The freezing point depression can be Determine by subtracting the value of the initial temperature from the final temperature. Therefore;

The freezing point depression = [ 1 - (-3)]° C = 4°C.

(b). The molality can be Determine by using the formula below;

Molality = the number of moles found in the solute/ solvent's weight(kg).

Molality = ( 11.1 / 58) × (1000)/ ( 90.4 - 11.1) = 2.4 M.

(c). The mass of acetone that was in the decanted solution = 11.1 g.

(d). The mass of water that was in the decanted solution = 89.01 g.

(e). 2.4 = x/ 58 × (1000/1000).

x = 2.4 × 58 = 139.2 g.

(f). The molar mass of acetone = (12) + (1 × 3) + 12 + 16 + 12 + (1 x 3) = 58 g/mol.

the dissolution of polymer can be thermodynamically described in 2 steps.