A solution is prepared by dissolving 4.9g of sucrose in 17g water. Calculate the boiling point, freezing point and osmotic pressure of this solution at 25 degree Celsius.
To calculate the boiling point, freezing point, and osmotic pressure of a solution, you need to use colligative properties. Colligative properties are dependent on the number of particles in a solution rather than the type of particles.
1. Boiling Point Elevation:
The boiling point elevation equation is given by:
ΔTb = Kb * m
Where:
ΔTb is the boiling point elevation
Kb is the molal boiling point elevation constant (0.512 °C kg/mol for water)
m is the molality of the solution (moles of solute/kg of solvent)
First, you need to calculate the molality (m) of the solution:
Molar mass of sucrose (C12H22O11) = 342.3 g/mol
Number of moles of sucrose = mass of sucrose / molar mass of sucrose
= 4.9 g / 342.3 g/mol
≈ 0.0143 mol
Molality (m) = moles of solute / kg of solvent
= 0.0143 mol / 0.017 kg
≈ 0.84 mol/kg
Now, calculate the boiling point elevation:
ΔTb = Kb * m
ΔTb = 0.512 °C kg/mol * 0.84 mol/kg
ΔTb ≈ 0.43 °C
The boiling point of water at 25 °C is 100 °C. Therefore, the boiling point of the solution would be:
Boiling point of solution = boiling point of water + ΔTb
Boiling point of solution = 100 °C + 0.43 °C
Boiling point of solution ≈ 100.43 °C
2. Freezing Point Depression:
The freezing point depression equation is given by:
ΔTf = Kf * m
Where:
ΔTf is the freezing point depression
Kf is the molal freezing point depression constant (-1.86 °C kg/mol for water)
Calculate the freezing point depression:
ΔTf = Kf * m
ΔTf = -1.86 °C kg/mol * 0.84 mol/kg
ΔTf ≈ -1.56 °C
The freezing point of water at 25 °C is 0 °C. Therefore, the freezing point of the solution would be:
Freezing point of solution = freezing point of water + ΔTf
Freezing point of solution = 0 °C - 1.56 °C
Freezing point of solution ≈ -1.56 °C
3. Osmotic Pressure:
The osmotic pressure equation is given by:
π = i * M * R * T
Where:
π is the osmotic pressure
i is the van't Hoff factor (the number of particles into which the solute dissociates in solution)
M is the molarity of the solution (moles of solute/L of solution)
R is the ideal gas constant (0.0821 L·atm/(mol·K))
T is the temperature in Kelvin (25 °C + 273.15)
Calculate the molarity (M) of the solution:
Moles of solute = mass of sucrose / molar mass of sucrose
= 4.9 g / 342.3 g/mol
≈ 0.0143 mol
Molarity (M) = moles of solute / L of solution
= 0.0143 mol / (0.017 kg + 0.017 L)
≈ 0.43 mol/L
Calculate the osmotic pressure:
π = i * M * R * T
Assuming sucrose does not dissociate (i = 1),
π = 1 * 0.43 mol/L * 0.0821 L·atm/(mol·K) * (25 °C + 273.15)
π ≈ 9.1 atm
Therefore, the boiling point of this solution at 25 °C is approximately 100.43 °C, the freezing point is approximately -1.56 °C, and the osmotic pressure is approximately 9.1 atm.
mols sucrose = grams/molar mass = 4.9/342 = ?
m = molality = mols sucrose/kg solvent = ?mols sucrose/0.017 kg = ?
dT = Kf*m. You have Kf for water and molality from above. Solve for delta T then normal freezing point - delta T = new freezing point. Kf is 1.86 for water.
dT = Kb*m
You have Kb for water (0.512), m from above, solve for delta T. Then normal boiling point for H2O + delta T = new boiling point.
osmotic pressure = MRT
You know T (remember to use kelvin), and R (universal gas constant). Convert molality (m) to molarity (M) and solve for osmotic pressure. For very dilute solutions m may be used interchangeably with M
Post your work if you get stuck.
Well, let me calculate that for you. But before I do, did you know that water boils at 100 degrees Celsius, freezes at 0 degrees Celsius, and osmotic pressure is like the pressure you feel when you have a million thoughts running through your head?
Now, back to your question. To calculate the boiling point elevation, we can use the equation:
ΔTb = Kb * m
Where:
ΔTb is the boiling point elevation,
Kb is the ebullioscopic constant for water (0.512 °C/m),
and m is the molality of the solution.
First, we need to find the molality, which is defined as the number of moles of solute per kilogram of solvent. Since we have 4.9g of sucrose and 17g of water, we can calculate the molality as follows:
molality = (mass of sucrose / molar mass of sucrose) / (mass of water / molar mass of water)
Now, the molar mass of sucrose is 342.3 g/mol, and the molar mass of water is 18 g/mol. Plug those values into the equation and divide by 1000 to convert grams to kilograms, and we get... drumroll, please!
molality = (4.9 g / 342.3 g/mol) / (17 g / 18 g/mol)
molality ≈ 0.00157 mol/kg
Now, let's calculate the boiling point elevation:
ΔTb = (0.512 °C/m) * (0.00157 mol/kg)
ΔTb ≈ 0.00080784 °C
So the boiling point of the solution would be slightly higher than 100 degrees Celsius. Maybe it'll reach a scorching 100.00080784 degrees Celsius!
Now, for the freezing point depression, we use the equation:
ΔTf = Kf * m
Where:
ΔTf is the freezing point depression,
Kf is the cryoscopic constant for water (1.86 °C/m),
and m is the molality of the solution (which we already calculated).
Using the same molality value, we can find the freezing point depression:
ΔTf = (1.86 °C/m) * (0.00157 mol/kg)
ΔTf ≈ 0.0029152 °C
So the freezing point of the solution would be slightly lower than 0 degrees Celsius. It's like a popsicle that's melting before you even take it out of the freezer!
Lastly, let's calculate the osmotic pressure. We can use the equation:
π = i * M * R * T
Where:
π is the osmotic pressure,
i is the van't Hoff factor (which is 1 for sucrose),
M is the molality of the solution,
R is the ideal gas constant (0.0821 L * atm / mol * K),
and T is the temperature in Kelvin (25 + 273.15 K).
Let's plug in the values and find out the osmotic pressure:
π = (1) * (0.00157 mol/kg) * (0.0821 L * atm / mol * K) * (25 + 273.15 K)
π ≈ 0.1060 atm
So the osmotic pressure of the solution would be approximately 0.1060 atmospheres. Just enough to make our balloon animals!
To calculate the boiling point, freezing point, and osmotic pressure of the solution, we can use the following formulas:
1. For boiling point elevation: ΔTb = Kb * m * i
2. For freezing point depression: ΔTf = Kf * m * i
3. For osmotic pressure: Π = M * R * T
Where:
- ΔTb is the boiling point elevation
- ΔTf is the freezing point depression
- Kb is the molal boiling point elevation constant (0.512 °C/m for water)
- Kf is the molal freezing point depression constant (1.86 °C/m for water)
- m is the molality of the solution (mol solute/kg solvent)
- i is the Van't Hoff factor, which represents the number of particles formed upon dissociation (1 for sucrose)
- Π is the osmotic pressure
- M is the molarity of the solution (mol solute/L solution)
- R is the ideal gas constant (0.0821 L⋅atm/(mol⋅K))
- T is the temperature in Kelvin (25 + 273.15 = 298.15 K)
Now let's calculate the results step by step:
1. Calculate the molality (m) of the solution:
molality (m) = moles of solute / mass of solvent (in kg)
Since the molar mass of sucrose (C12H22O11) is approximately 342.3 g/mol:
molality (m) = 4.9 g / (17 g / 1000) = 4.9 g / 0.017 kg = 288.23 mol/kg
2. Calculate the boiling point elevation (ΔTb):
ΔTb = Kb * m * i
Kb for water is 0.512 °C/m:
ΔTb = 0.512 °C/m * 288.23 mol/kg * 1 = 147.66 °C
The boiling point elevation is 147.66 °C.
3. Calculate the freezing point depression (ΔTf):
ΔTf = Kf * m * i
Kf for water is 1.86 °C/m:
ΔTf = 1.86 °C/m * 288.23 mol/kg * 1 = 536.68 °C
The freezing point depression is 536.68 °C.
4. Calculate the osmotic pressure (Π):
Π = M * R * T
First, we need to calculate the molarity (M) of the solution:
Molarity (M) = moles of solute / volume of solution
To find the volume of the solution, we need to convert the mass of water to volume. The density of water is approximately 1 g/mL:
Volume of water = mass of water / density of water = 17 g / 1 g/mL = 17 mL = 0.017 L
Now, let's calculate the molarity:
M = 4.9 g / (0.017 L / 1000) = 4.9 g / 0.017 L = 288.23 mol/L
Now we can calculate the osmotic pressure:
Π = 288.23 mol/L * 0.0821 L⋅atm/(mol⋅K) * 298.15 K = 6943.85 atm
The osmotic pressure is 6943.85 atm.