Glycerol (a non-electrolyte that can be used
as antifreeze and coolant in a car radiator)
has a density of 1.261 g/mL and a molecular
weight of 92 g/mol. If 111 mL of glycerol
are mixed with 5.1 kg of water, by how much
would the boiling point of the solution be
raised above that of pure water? Kb for water
= 0.512◦C/m.
Answer in units of ◦C.
To find the boiling point elevation of the glycerol-water solution, we can use the formula:
ΔTb = Kb * molality
Where:
ΔTb is the boiling point elevation
Kb is the boiling point elevation constant for water (given as 0.512°C/m)
molality is the concentration of the solution expressed as moles of solute per kilogram of solvent
First, let's calculate the number of moles of glycerol and the molality of the solution:
Step 1: Calculate the number of moles of glycerol
Given:
- Density of glycerol = 1.261 g/mL
- Molecular weight of glycerol = 92 g/mol
- Volume of glycerol = 111 mL
We can calculate the mass of glycerol:
Mass of glycerol = Density * Volume = 1.261 g/mL * 111 mL = 140 g
Now, we can calculate the number of moles:
Moles of glycerol = Mass / Molecular weight = 140 g / 92 g/mol ≈ 1.522 moles
Step 2: Calculate the molality of the solution
Given:
- Mass of water = 5.1 kg
We need to convert the mass of water to grams:
Mass of water = 5.1 kg * 1000 g/kg = 5100 g
Now, we can calculate the molality:
Molality = Moles of glycerol / Mass of water (in kg)
= 1.522 moles / 5.1 kg
≈ 0.298 mol/kg
Step 3: Calculate the boiling point elevation
Now that we have the molality, we can use the formula mentioned earlier:
ΔTb = Kb * molality
ΔTb = 0.512°C/m * 0.298 mol/kg
Finally, we can calculate ΔTb:
ΔTb = 0.153°C
Therefore, the boiling point of the solution would be raised by approximately 0.153°C above that of pure water.