Question

A solution is made by dissolving 23.5 grams of glucose (C6H12O6) in .245 kilograms of water. If the molal freezing point constant for water (Kf) is -1.86 °C/m, what is the resulting Δtf of the solution? Show all the steps taken to solve this problem.

Answer 1

@Photon336

Answer 2

@dan815

Answer 3

@abb0t

Answer 4

@mathmate

Answer 5

@SyedMohammed98

Answer 6

@zepdrix

Answer 7

@paki

Answer 8

@sammixboo

Answer 9

@Ciarán95

Answer 10

@TheRaggedyDoctor

Answer 11

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Answer 12

Hold on until tara or photon come online. I haven't still learned these stuff. Sorry I couldn't help u@@@@@

Answer 13

@tywower

do you know this formula? \[\Delta T = ikbm \]

delta T is the temperature changed caused by adding a certain amount of solute in the solvent.

m = molality moles of solute per kilogram of solvent.
i = Van Hoff factor
Kb = our constant.

Answer 14

convert grams of glucose to moles of glucose

\[23.5g x (1mol/180g) = 23.5/180 = 0.130 \]

they gave us the amount of kilograms of water which is = 0.245 kg

now we calculate the total molality:

\[molality = moles/kilogram of solvent = 0.130moles/0.245kg = 0.53 m \]

we put all of this together:

kf = -1.86 C/m

\[-1.86 C/m *0.53 m*1 = -0.99 \]

we realize that the temperature changed by a factor of -0.99 or 1 and the freezing point of water we know is 0 degrees celsius so our new temperature = -1 degree celsius. so what this means is that by adding a certain amount of solute the freezing point went down by 1 degree celsius.