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
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.
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 = 100º + 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.
To determine the boiling point, we can use the concept of colligative properties, specifically the boiling point elevation. The boiling point elevation is directly proportional to the molality of the solute particles.
First, we need to calculate the molality of the solute particles (ethylene, C2H6O2). Molality is defined as the number of moles of solute divided by the mass of the solvent (in kilograms). The molar mass of ethylene (C2H6O2) is 46.07 g/mol.
Molality (m) = moles of solute / mass of solvent (in kg)
Given:
- Moles of ethylene (C2H6O2) = 3.15 moles
- Mass of water (H2O) = 250.0 g
First, convert the mass of water to kilograms:
Mass of water (H2O) = 250.0 g = 0.2500 kg
Now, calculate the molality:
Molality (m) = 3.15 moles / 0.2500 kg = 12.60 m
Next, we can use the boiling point elevation equation to calculate the boiling point elevation (∆Tb):
∆Tb = Kb * m
Where Kb is the cryoscopic constant and depends on the solvent. For water, Kb = 0.512 °C/m.
∆Tb = 0.512 °C/m * 12.60 m = 6.4512 °C
Lastly, we add the boiling point elevation (∆Tb) to the normal boiling point of water, which is 100 °C:
Boiling point = Normal boiling point + ∆Tb
Boiling point = 100 °C + 6.4512 °C = 106.4512 °C
Rounded to the nearest hundredth, the boiling point is 106.45 °C.
Therefore, the correct answer is option c) 106.45 °C.