delta T = Kb*molality
You know delta T (0.500C) and Kb, solve for m.
m = mols solute/kg solvent.
You know m and kg solvent, solve for mols
Then mols solute = grams/molar mass. You know mols and molar mass, solve for grams.
How many grams of cholesterol, C27H46O (386.6 g/mol), must be dissolved in 208.0 grams of diethyl ether to raise the boiling point by 0.5000C ?
You know delta T (0.500C) and Kb, solve for m.
m = mols solute/kg solvent.
You know m and kg solvent, solve for mols
Then mols solute = grams/molar mass. You know mols and molar mass, solve for grams.
ΔTb = Kb * m * i
Where:
ΔTb is the boiling point elevation,
Kb is the molal boiling point elevation constant,
m is the molality of the solute (in this case, cholesterol),
and i is the van't Hoff factor.
We can rearrange the equation to solve for m:
m = ΔTb / (Kb * i)
Given values:
ΔTb = 0.5000C
Kb of diethyl ether = 2.02C/m
i for cholesterol is assumed to be 1 since it is a nonelectrolyte.
Plugging in the values:
m = 0.5000C / (2.02C/m * 1) = 0.2475 m
Now, we need to convert molality to moles of cholesterol per kg of diethyl ether.
0.2475 m = 0.2475 moles of cholesterol / 1 kg of diethyl ether
To convert moles to grams, we need to multiply by the molar mass of cholesterol:
0.2475 moles * 386.6 g/mol = 95.52 grams
Therefore, approximately 95.52 grams of cholesterol must be dissolved in 208.0 grams of diethyl ether to raise the boiling point by 0.5000C.