A solution is prepared by dissolving 4.9 grams of sucrose in 17gram of water.
At 25 degree Celsius find the boiling point, freezing point and osmotic pressure of this solution.
Confused? At 25 C how can it have a boiling point? You know it must be over 100 C.
To find the boiling point, freezing point, and osmotic pressure of a solution, we need to use the formula:
ΔT = K * m
Where:
- ΔT is the change in temperature
- K is the molal boiling point/freezing point constant for the solvent (water)
- m is the molality of the solution
First, let's calculate the molality (m) of the solution:
Molality (m) = moles of solute / mass of solvent (in kg)
Given:
- Mass of solute (sucrose) = 4.9 grams
- Mass of solvent (water) = 17 grams
We need to convert the masses to kg:
- Mass of solute (sucrose) = 4.9 grams = 0.0049 kg
- Mass of solvent (water) = 17 grams = 0.017 kg
Now, let's calculate the molality (m):
Molality (m) = 0.0049 kg / 0.017 kg
= 0.288 mol/kg
Next, let's calculate the boiling point elevation using the boiling point constant for water (K_b):
K_b for water = 0.512 °C/m
Boiling point elevation (ΔT_b) = K_b * m
= 0.512 °C/m * 0.288 mol/kg
= 0.147456 °C
To find the boiling point of the solution, we add the boiling point elevation to the normal boiling point of water:
Boiling point of water = 100°C
Boiling point of the solution = 100°C + 0.147456°C
= 100.147456°C
Now, let's calculate the freezing point depression using the freezing point constant for water (K_f):
K_f for water = 1.86 °C/m
Freezing point depression (ΔT_f) = K_f * m
= 1.86 °C/m * 0.288 mol/kg
= 0.53568 °C
To find the freezing point of the solution, we subtract the freezing point depression from the normal freezing point of water:
Freezing point of water = 0°C
Freezing point of the solution = 0°C - 0.53568°C
= -0.53568°C
Finally, let's calculate the osmotic pressure (π) using the formula:
π = i * M * R * T
Where:
- i is the van't Hoff factor (number of particles after dissociation)
- M is the molarity of the solution
- R is the ideal gas constant (0.08206 L.atm/mol.K)
- T is the temperature in Kelvin (25°C = 298K)
In this case, sugar does not dissociate, so i = 1.
Molarity (M) = moles of solute / volume of solvent (in liters)
Given:
- Mass of solute (sucrose) = 4.9 grams
- Molecular weight of sucrose (C12H22O11) = 342 grams/mol
- Volume of solvent (water) = 17 grams = 0.017 L
First, let's calculate the moles of sucrose:
Moles of sucrose = mass of sucrose / molecular weight of sucrose
= 4.9 grams / 342 g/mol
= 0.0143 mol
Next, let's calculate the molarity (M):
Molarity (M) = 0.0143 mol / 0.017 L
= 0.8412 M
Finally, let's calculate the osmotic pressure (π):
π = 1 * (0.8412 M) * (0.08206 L.atm/mol.K) * (298 K)
= 20.503 atm
So, the boiling point of the solution is 100.147456°C, the freezing point of the solution is -0.53568°C, and the osmotic pressure of the solution is 20.503 atm at 25°C.
To find the boiling point, freezing point, and osmotic pressure of a solution, we need to use some formulas and known values. Here's how you can calculate each of these properties:
1. Boiling Point Elevation:
The boiling point elevation (ΔTb) of a solution can be calculated using the formula:
ΔTb = Kb * m
Where Kb is the molal boiling point constant for the solvent and m is the molality of the solution.
For water, the molal boiling point constant (Kb) is approximately 0.52 °C·kg/mol.
First, calculate the molality (m) of the solution:
m = moles of solute / mass of solvent (in kg)
Given that the mass of sucrose is 4.9 grams and the mass of water is 17 grams:
Moles of sucrose = mass of sucrose / molar mass of sucrose
Molar mass of sucrose (C12H22O11) = (12 * 12.01 g/mol) + (22 * 1.01 g/mol) + (11 * 16.00 g/mol) = 342.3 g/mol
Moles of sucrose = 4.9 g / 342.3 g/mol
Mass of water = 17 g = 17/1000 kg (conversion from grams to kg)
Now, plug in the values into the formula to find ΔTb:
ΔTb = 0.52 °C·kg/mol * (moles of sucrose / mass of water in kg)
2. Freezing Point Depression:
The freezing point depression (ΔTf) of a solution can be calculated using the formula:
ΔTf = Kf * m
Where Kf is the molal freezing point constant for the solvent and m is the molality of the solution.
For water, the molal freezing point constant (Kf) is approximately 1.86 °C·kg/mol.
Using the same formula as above, calculate the molality (m) of the solution and then plug it into the formula to find ΔTf.
3. Osmotic Pressure:
The osmotic pressure (π) of a solution can be calculated using the formula:
π = i * M * R * T
Where i is the van't Hoff factor, M is the molarity of the solution, R is the ideal gas constant, and T is the temperature in Kelvin.
The van't Hoff factor (i) is the number of particles the solute dissociates into when it dissolves. For sucrose (C12H22O11), it does not dissociate, so i would be 1.
To find M, first calculate the moles of sucrose (as done above) and then divide it by the volume of the solution in liters.
M = moles of sucrose / volume of solution (in liters)
R is the ideal gas constant, which is approximately 0.0821 L·atm/(mol·K).
Convert the temperature to Kelvin:
T (Kelvin) = 25 + 273.15
Now, plug in the values into the formula to find osmotic pressure (π).
By performing the above calculations, you will be able to find the boiling point elevation, freezing point depression, and osmotic pressure of the given solution.