1. An aqueous solution containing 35.0% by mass of an unknown covalent compund i
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Question
1. An aqueous solution containing 35.0% by mass of an unknown covalent compund in water was found to have a freezing point of -3.5 deg C. Calculate the molar mass of the unknown compund.
2. A 11.6 M aqueaous solution of ethylene glycol (C2H6O2) has the capability of lowering the freezing point of water to -45 deg C! The density of the solution is 1.09 g/mL and the molar mass of ethylene glycol is 62.07 g/mol.
a) what is the molality of ethylene glycol solution?
b) what is the mole fraction of the ethylene glycol solution?
3. A mixture of benzene and methylbenzene has a normal boiling point of 398K. Given the information below:
Substance Vapor Pressure at 398K Molar Mass
Benzene 1520 torr 78.11 g/mole
Methylbenzene 570 torr 92.14 g/mole
a) what is the mole fraction of benzene in the liquid (solution)?
b) what is the mole fraction of benzene in the vapor?
SHOW ALL WORK FOR POINTS!!!
Explanation / Answer
Depression in freezing point = kf.m
=> 3.5 = 1.853*m [m is the molality of solution]
=>m = 1.889 molal
This aqueous solution has (35/65)kg solute if the weight of solvent is 1kg
number of moles of solute = 1.889 if the solution has 1kg solvent.
=>1.889 = (35/65)/M => M = 35/(65*1.889) kg = 285.08 g/mol
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Delta(Tf) = kf.m
=> m = 45/1.853 molal = 24.285 molal
Molality of solution = 24.285 molal
concentration of solution = 11.6 M
11.6 moles of ethyleneglycol per liter solution.
Wt. of glycol in liter solution = 11.6*62.07g = 720 g
Weight of 1 liter solution = 1.09*1000 = 1090g
Weight of water = 1090-720=370g
Moles of water = 370/18 = 20.56 moles
Mole fraction of ethylene glycol = 11.6*(11.6+20.56) = 0.361
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Benzene and methylbenzene(toluene) form an ideal solution that follows raoult's law.
Benzene - Boiling point 80C
kb = 2.53C/m
Elevation in boiling point of mixture = 45C
Delta(Tb) = kb.m
=> m = 45/2.53 = 17.79
=> 17.79 moles of methylbenzene in 1kg benzene
1 kg of benzene = 12.80 moles [MW of benzene = 78.11]
Mixture 12.8 mol benzene : 17.79 mol methyl benzene
Mole fraction of benzene = 12.8/(12.8+17.79) = 0.418
Mole fraction of methylbenzene = 1-0.418 = 0.582
After the solution achieves liquid-vapor equilibrium,
Vapor pressure due to benzene = 0.418*1520 torr = 635.56torr
Vapor pressure due to methyl benzene = 0.582*570 = 331.74
Mole fraction of benzene in vapor phase = 635.56/(635.56+331.74) = 0.657
Mole fraction of benzene in liquid phase = 1-0.657 = 0.343
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