The boiling point of pure water in Winter Park, CO ( elev. 9000 ft)
is 94 oC. What is the boiling point of a solution containing 11.3 g
of glucose (180 g/mol) in 55 mL of water in Winter Park? Kb for
water = 0.512°C/m.
To determine the boiling point of a solution, you can use the equation:
ΔT = Kb × m
Where:
ΔT is the change in boiling point
Kb is the boiling point elevation constant for water
m is the molality of the solution
First, let's calculate the molality (m) of the solution.
Step 1: Convert the mass of glucose from grams to moles.
Given mass of glucose = 11.3 g
Molar mass of glucose = 180 g/mol
moles of glucose = mass / molar mass
moles of glucose = 11.3 g / 180 g/mol
moles of glucose ≈ 0.063 moles
Step 2: Calculate the mass of water using its density.
Given volume of water = 55 mL
Density of water = 1 g/mL
mass of water = volume × density
mass of water = 55 mL × 1 g/mL
mass of water = 55 g
Step 3: Calculate the molality of the solution.
molality (m) = moles of solute / mass of solvent (in kg)
mass of solvent = mass of water / 1000 (to convert from grams to kilograms)
mass of solvent = 55 g / 1000
mass of solvent = 0.055 kg
molality (m) = 0.063 moles / 0.055 kg
molality (m) ≈ 1.145 mol/kg
Now that we have the molality of the solution, let's calculate the change in boiling point (ΔT).
Given Kb = 0.512 °C/m
ΔT = Kb × m
ΔT = 0.512 °C/m × 1.145 mol/kg
ΔT ≈ 0.586 °C
The change in boiling point (ΔT) is approximately 0.586 °C.
To find the boiling point of the solution, you need to add the change in boiling point to the boiling point of pure water.
Boiling point of pure water in Winter Park, CO = 94 °C
Boiling point of the solution = Boiling point of pure water + ΔT
Boiling point of the solution = 94 °C + 0.586 °C
Boiling point of the solution ≈ 94.586 °C
Therefore, the boiling point of the solution containing 11.3 g of glucose in 55 mL of water in Winter Park, CO (elevation 9000 ft) is approximately 94.586 °C.
To find the boiling point of a solution, we can use the formula:
ΔTb = Kb * m
Where ΔTb is the boiling point elevation, Kb is the molal boiling point constant for the solvent, and m is the molality of the solution.
First, let's calculate the molality of the solution:
Moles of glucose (C6H12O6) = mass / molar mass
Moles of glucose = 11.3 g / 180 g/mol
Moles of glucose = 0.0628 mol
Molality (m) = moles of solute / mass of solvent (in kg)
Mass of solvent = volume of solvent (in L) x density of water (1 g/mL)
Volume of solvent = 55 mL = 55/1000 L = 0.055 L
Mass of solvent = 0.055 L x 1 g/mL = 0.055 kg
Molality (m) = 0.0628 mol / 0.055 kg
Molality = 1.142 mol/kg
Now, we can calculate the boiling point elevation:
ΔTb = Kb * m
ΔTb = 0.512 °C/m * 1.142 mol/kg
ΔTb = 0.585 °C
Finally, we can find the boiling point of the solution:
Boiling point = boiling point of pure water + ΔTb
Boiling point = 94 °C + 0.585 °C
Boiling point = 94.585 °C
So, the boiling point of the solution containing 11.3 g of glucose in 55 mL of water in Winter Park, CO is approximately 94.585 °C.
mols glucose = grams/molar mass = ?
m = mols glucose/kg solvent
delta T = Kb*m
Add delta T to 94 to find the new boiling point.