Hello, I just need some help with the Pre-Lab exercises. I tried to to do them o

Question

Hello,

I just need some help with the Pre-Lab exercises. I tried to to do them on my own but I keep getting confused on what to do. Any help would be appreciated. :)

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 deg C on her thermometer. When she added 11.1 g of her sample into the mixture, the temperature, after thorough stirring, fell to -3.0 deg 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 the acetone

C. How much acetone was in the decanted solution?

D. How much water was 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?

Please show step by step. I'm sure I'm just over thinking everything but I just need some help. Thank you! :)

Explanation / Answer

a. what was the freezing point depression

delta t = 1- (-3) = 4 degrees C

b. what was the molality of the acetone

we have 11.1 g of the acetone in 90.4g of solution
mass of water is 79.3g
we have 11.1 g of acetone in 79.3 g of water
mass in 1000 g = 11.1 x 1000 / 79.3 = 140 g / kg

c. how much acetone was in the decanted solution

11.1 g

d. how much water was in the decanted solution

79.3 g

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

140 g of acetone / kg of water

f. what did she find the molar mass of acetone, assuming she made the calculation properly

delta t = m x kf x 1 ; here we have 1 particle so i = 1; kf for water is 1.86

m = 4/ 1.86 = 2.151 moles of acetone / kg of water

so in 1 kg of water we have 2.410 moles = 140g

mass of 1 mole = 140 / 2.410 = 58.09 g / mole .... excellent result

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