calculate the freezing and boiling points of each of the following solutions.
(a) 0.37 m glucose in ethanol
freezing point ______I got -1.2e2 but its wrong_______°C
boiling point _________I got 1.1e2 but its wrong:(____°C
(b) 15.0 g of decane, C10H22, in 37.9 g CHCl3
freezing point _____________°C
boiling point _____________°C
Thanks:)
To calculate the freezing and boiling points of a solution, we need to consider the concept of freezing point depression and boiling point elevation. These phenomena occur when a solute is added to a solvent and affect the freezing and boiling points of the resulting solution.
(a) For the solution of 0.37 m glucose in ethanol:
To calculate freezing point depression, we can use the equation:
ΔTf = Kf * m
Where ΔTf is the freezing point depression, Kf is the cryoscopic constant of the solvent, and m is the molality of the solute.
For ethanol, the cryoscopic constant (Kf) is approximately 1.99 °C·kg/mol.
First, we need to calculate the molality (m) of glucose in ethanol:
Molality (m) = moles of solute / mass of solvent (in kg)
In this case, assume we have 1 L of ethanol, which has a density of 0.789 g/mL. Thus, 1 L of ethanol has a mass of 789 g or 0.789 kg.
The molar mass of glucose (C6H12O6) is 180.16 g/mol, so the number of moles of glucose is:
0.37 mol/L * 1 L = 0.37 moles
Using the mass of solvent, we can calculate the molality of glucose in ethanol:
m = 0.37 mol / 0.789 kg = 0.469 mol/kg
Now, we can calculate the freezing point depression:
ΔTf = 1.99 °C·kg/mol * 0.469 mol/kg = 0.933 °C
To find the freezing point of the solution, subtract the freezing point depression from the freezing point of the pure solvent:
Freezing point = Freezing point of pure solvent - ΔTf
Since the freezing point of ethanol is approximately -114 °C, the freezing point of the solution is:
Freezing point = -114 °C - 0.933 °C = -114.933 °C
For the boiling point elevation, we can use the equation:
ΔTb = Kb * m
Where ΔTb is the boiling point elevation, Kb is the ebullioscopic constant of the solvent, and m is the molality of the solute.
The ebullioscopic constant (Kb) for ethanol is approximately 1.22 °C·kg/mol.
We can use the same molality (m) calculated above, which is 0.469 mol/kg.
Now, we can calculate the boiling point elevation:
ΔTb = 1.22 °C·kg/mol * 0.469 mol/kg = 0.572 °C
To find the boiling point of the solution, add the boiling point elevation to the boiling point of the pure solvent:
Boiling point = Boiling point of pure solvent + ΔTb
Since the boiling point of ethanol is approximately 78 °C, the boiling point of the solution is:
Boiling point = 78 °C + 0.572 °C = 78.572 °C
(b) For the solution of 15.0 g of decane (C10H22) in 37.9 g of CHCl3:
To calculate the freezing and boiling points, we can use the same approach as in part (a).
First, calculate the molality (m) of decane in CHCl3:
Molar mass of decane (C10H22) = 142.28 g/mol
Number of moles of decane = 15.0 g / 142.28 g/mol = 0.1054 mol
Mass of CHCl3 = 37.9 g
Molality (m) = moles of solute / mass of solvent (in kg)
m = 0.1054 mol / 0.0379 kg = 2.777 mol/kg
Next, calculate the freezing point depression:
ΔTf = Kf * m
The cryoscopic constant (Kf) for CHCl3 is approximately 4.68 °C·kg/mol.
ΔTf = 4.68 °C·kg/mol * 2.777 mol/kg = 12.99 °C
To find the freezing point of the solution:
Freezing point = Freezing point of pure solvent - ΔTf
Since the freezing point of CHCl3 is approximately -63.5 °C, the freezing point of the solution is:
Freezing point = -63.5 °C - 12.99 °C = -76.49 °C
Now, calculate the boiling point elevation:
ΔTb = Kb * m
The ebullioscopic constant (Kb) for CHCl3 is approximately 3.63 °C·kg/mol.
ΔTb = 3.63 °C·kg/mol * 2.777 mol/kg = 10.09 °C
To find the boiling point of the solution:
Boiling point = Boiling point of pure solvent + ΔTb
Since the boiling point of CHCl3 is approximately 61.15 °C, the boiling point of the solution is:
Boiling point = 61.15 °C + 10.09 °C = 71.24 °C
Therefore, the freezing point of the 0.37 m glucose in ethanol solution is approximately -114.933 °C, and the boiling point is approximately 78.572 °C. The freezing point of the 15.0 g of decane in 37.9 g CHCl3 solution is approximately -76.49 °C, and the boiling point is approximately 71.24 °C.