A solution prepared at 20 degrees celsius by dissolving 7.44g of an unknown hydrocarbon in 75g benzene (C6H6, 78.112 g/mol, 0.8765 g/cm^3) is found to have a vapor pressure of 376 mmHg at 60 degrees celsius. The vapor pressure of pure benzene at 60 degrees celsius is 400 mmHg. Determine the molar mass of the unknown compound.
Solution :-
Mass of unknown hydrocarbon = 7.44 g
Mass of benzene = 75 g
Molar mass of benzene = 78.112 g per mol
Vapor pressure of benzene = 400 mmHg
Vapor pressure of solution = 376 mmHg
Molar mass of unknown hydrocarbon =?
Using the Raoults law formula we can calculate the mole fraction of the benzene.
Psolution = (χsolvent) (P°solvent)
Where , Psolution = vapor pressure of solution , (χsolvent) = mole fraction of solvent , = vapor pressure of pure solvent
Now lets put the values in the formula
376 mmHg = (χsolvent) * 400 mmHg
376 mmHg / 400 mmHg = (χsolvent)
0.94 = (χsolvent)
Therefore mole fraction of the solvent = 0.94
Mole fraction of the solute = 1- 0.94 = 0.06
Now lets calculate the moles of benzene using its given mass
Moles = mass / molar mass
Moles of Benezene =75 g /78.112 g per mol = 0.96016 mol benzene
Now use the mole of the benzene and mole fraction of benzene to determine the moles of the unknown hydrocarbon
0.94 = 0.96016 / (0.96016+x)
X is nothing but the moles of the unknown hydrocarbon
0.94 * 0.96016+x =0.96016
0.90255 +0.94 x =0.96016
0.94 x = 0.05851
X = 0.05851/0.94
X =0.062244 mol unknown hydrocarbon
Now lets use this moles of the unknown hydrocarbon to calculate its molar mass
Molar mass = mass / moles
Molar mass of unknown hydrocarbon = 7.44 g / 0.062244 mol
= 119.53 g per mol
Therefore molar mass of the unknown hydrocarbon = 119.53 g / mol
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