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Finding molal freezing point

A solution prepared by dissolving 3.00g of ascorbic acid (vitamin C, C6H8O6 ) in 50.0g of acetic acid has a freezing point that is depressed by ΔT = 1.33°C below that of pure acetic acid.
Part A
What is the value of the molal freezing-point-depression constant for acetic acid?
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1 Answer

Molecular weight (or mass) of ascorbic acid is 176.12 g/mole
 
(3g)/(176.12 g/mole) = 0.017 mole
 
concentration = (0.017 mole)/(50 g) = (0.017 mole)/[(50 g)/(1000 g/kg)] = 0.341 moles/kg
 
Concentration in terms of molality = 0.341 = cm
 
ΔT = (Kf)(cm) = 1.33 = (Kf)(0.341 moles/kg)
 
Kf = (1.33)/0.341) = 3.90 °C m-1