Calculate the boiling point elevation of a solution containing 22.8 g of glucose (C6H12O6) dissolved in 672.0 g of water. Calculate the freezing point depression for the same solution.
delta T = Kb*molality
molality = moles/kg solvent
moles = g/molar mass
[convert 22.8 g glucose to moles using equation 3. Convert moles glucose to molality using equation 2. Then calculate delta T using equation 1.]
Then delta T = Kf*m
for the freezing point section. The same moles and molality may be used in this part.
To calculate the boiling point elevation and freezing point depression of a solution, we need to use the following equations:
For boiling point elevation:
ΔTb = Kb * m
Where:
ΔTb = boiling point elevation (in °C)
Kb = ebullioscopic constant for the solvent (in °C/m)
m = molality of the solute (in mol/kg)
For freezing point depression:
ΔTf = Kf * m
Where:
ΔTf = freezing point depression (in °C)
Kf = cryoscopic constant for the solvent (in °C/m)
m = molality of the solute (in mol/kg)
First, we need to calculate the molality of the solute (glucose) in the solution.
Step 1: Calculate the moles of glucose
Molar mass of glucose (C6H12O6) = (6 * atomic mass of C) + (12 * atomic mass of H) + (6 * atomic mass of O)
= (6 * 12.01 g/mol) + (12 * 1.01 g/mol) + (6 * 16.00 g/mol)
= 180.18 g/mol
Moles of glucose = mass of glucose / molar mass of glucose
= 22.8 g / 180.18 g/mol
≈ 0.1264 mol
Step 2: Calculate the molality of the solute
Molality (m) = moles of solute / mass of solvent (in kg)
Mass of water = 672.0 g = 0.6720 kg
Molality (m) = 0.1264 mol / 0.6720 kg
≈ 0.1880 mol/kg
Now we can calculate the boiling point elevation and freezing point depression.
Step 3: Calculate the boiling point elevation
For water, Kb = 0.512 °C/m
ΔTb = Kb * m
= 0.512 °C/m * 0.1880 mol/kg
≈ 0.0963 °C
The boiling point of water is 100 °C, so the boiling point elevation will be:
Boiling point elevation = 100 °C + 0.0963 °C
≈ 100.0963 °C
Step 4: Calculate the freezing point depression
For water, Kf = 1.86 °C/m
ΔTf = Kf * m
= 1.86 °C/m * 0.1880 mol/kg
≈ 0.3497 °C
The freezing point of water is 0 °C, so the freezing point depression will be:
Freezing point depression = 0 °C - 0.3497 °C
≈ -0.3497 °C
Therefore, the boiling point elevation for the solution is approximately 0.0963 °C and the freezing point depression is approximately -0.3497 °C.
To calculate the boiling point elevation and freezing point depression of a solution, we need to use the molal concentration of the solute and the respective constants for the solvent.
First, let's calculate the molal concentration of glucose (C6H12O6) in the solution:
1. Calculate the moles of glucose:
Molar mass of glucose (C6H12O6) = 6*(12.01 g/mol) + 12*(1.01 g/mol) + 6*(16.00 g/mol) = 180.18 g/mol
Moles of glucose = Mass of glucose / Molar mass of glucose = 22.8 g / 180.18 g/mol
2. Calculate the molal concentration:
Molal concentration (m) = Moles of solute / Mass of solvent (in kg)
Mass of solvent = 672.0 g = 672.0 g / 1000 g/kg = 0.672 kg
Molal concentration = Moles of glucose / Mass of solvent
Now, we need to use the respective constants for water:
1. Boiling point elevation constant (Kb) for water: Kb = 0.512 °C/m
2. Freezing point depression constant (Kf) for water: Kf = 1.86 °C/m
3. Calculate the boiling point elevation:
Boiling point elevation (ΔTb) = Kb * m
4. Calculate the freezing point depression:
Freezing point depression (ΔTf) = Kf * m
Now, let's plug in the values:
1. Boiling point elevation:
ΔTb = 0.512 °C/m * Molal concentration (m) of glucose
2. Freezing point depression:
ΔTf = 1.86 °C/m * Molal concentration (m) of glucose
Note: The molal concentration is the same for both boiling point elevation and freezing point depression, as it depends only on the number of moles of solute and the mass of the solvent.
Now you can calculate the boiling point elevation and freezing point depression by plugging in the molal concentration value you have obtained.