Asked by brandi r
what is the boiling point that contains 3.15 moles of ethylene C2H6O2 and 250.0g of H2O? answer should end in degree's celcius
a) 1.03 C
b) 93.55 C
c) 106.45 C
d) 6.45 C
e) 103.44 C
a) 1.03 C
b) 93.55 C
c) 106.45 C
d) 6.45 C
e) 103.44 C
Answers
Answered by
DrBob222
molality = mols/kg solvent.
Solve for molality.
delta T = Kb*molality
Solve for delta T.
Add delta T + normal boiling point to obtain new boiling point.
Solve for molality.
delta T = Kb*molality
Solve for delta T.
Add delta T + normal boiling point to obtain new boiling point.
Answered by
GK
3.15 mol / 0.250 kg = 12.6 mol/kg or 12.6 molal
The boiling point of a solution is
Tb = Tb(H2O) + deltaT = 100ยบ + deltaT
deltaT = Kb*m = (0.512deg.kg/mol)(12.6mol/kg)
NOTE: C2H6O2 is not ethylene. It may be
CH3-O-O-CH3 whose solubility could not be 12.6molal. It is a bad question. The solution given gives the most likely *expected* answer.
The boiling point of a solution is
Tb = Tb(H2O) + deltaT = 100ยบ + deltaT
deltaT = Kb*m = (0.512deg.kg/mol)(12.6mol/kg)
NOTE: C2H6O2 is not ethylene. It may be
CH3-O-O-CH3 whose solubility could not be 12.6molal. It is a bad question. The solution given gives the most likely *expected* answer.
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