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Question:
Nicotine, extracted from tobacco leaves, is a pale yellow oil that is completely miscible with water at temperatures below 60 oC. What is the molality of nicotine in an aqueos solution that begins to freeze at -0.450 o C? If the solution is obtained by dissolving 1.921 g of nicotine in 48.92 g of water, what is the molar mass of nicotine?
Answer:
We need to use the equation for freezing point depression:
DTf = Kf m
For our problem:
DTf = freezing point depression
Kf = freezing point depression constant
m = molality (mol solute / kg solution)
Rearrange the equation to solve for molality (m):
m = 0.450 oC / 1.86 oC kg water (mol solute)-1
m = 0.242 mol solute / kg water
Now, use the definition of molality to determine the molecular weight of the substance:
m = (mol solute / kg solution)
Where:
mol solute = m / MW (mass of solute / molecular weight)
Rearrange the equation to solve for molecular weight (MW):
MW = 1.921 g / (0.04892 kg water * 0.242 mol solute / kg water)
MW = 162 g/mol
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