The boiling point of diethyl ether CH3CH2OCH2CH3 is 34.500C at 1 atmosphere. A nonvolatile, nonelectrolyte that dissolves in diethyl ether is cholesterol .
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 ?
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.
To calculate the grams of cholesterol that must be dissolved in diethyl ether, we need to use the equation for boiling point elevation:
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.