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