what is the freezing point of a solution of 12.0 g of CCl4 dissolved in 750.0g of benzene? the freezing point of benzene is 5.48 degrees C ; Kf is 5.12 degrees c/m.

Answer 1

convert 12.0 g CCl4 to moles. moles = grams/molar mass

molality = moles/kg solvent
solve for molality

delta T = Kf*m.
Solve for delta T and subtract from the normal freezing point of benzene.

Answer 2

4.948

Answer 3

1. convert 12.0 g CCl4 to moles. moles = grams / molar mass :

12.0 g CCl4 / 153.812 (g/mol) = 0.078 mol CCl4
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2. find the concentration (expressed in molality) of the solution. molality (m) = moles solute / kg solvent :

m = 0.078 mol CCl4 / 750.0 kg C6H6 [AKA benzene]

m = 0.104
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3. find the freezing point depression (refers to ΔTf) of the solution when compared to the solvent. ΔTf = Kf • m :

ΔTf = 5.12 ºC/m • 0.014m

ΔTf = 0.532ºC
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4. FINALLY, find the freezing point of the solution. Tf (the freezing point) of the solution is equal to Tf (the freezing point) of the solvent [benzene AKA C6H6] minus ΔTf :

5.48ºC - 0.532ºC =

4.948ºC
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II FINAL SOLUTION