Estimate the freezing point of 150 cm3 of water sweetened with 7.5g of sucrose
(C12H11O22). (Kf for water is 1.86 K kg mol-1; density of water is 1gcm-3)
moles sucrose = grams/molar mass. Solve for moles.
molality = moles sucrose/kg solvent. Solve for molality.
delta T = Kf*molality.
solve for delta T and from there the freezing point.
6.11*10^-5 K
To estimate the freezing point of the water sweetened with sucrose, we can use the formula:
ΔTf = Kf * m
Where:
ΔTf is the change in freezing point temperature
Kf is the cryoscopic constant for water (1.86 K kg mol-1)
m is the molality of the solution in mol/kg
To find the molality of the solution, we need to determine the number of moles of sucrose and the mass of water.
First, let's calculate the number of moles of sucrose:
Given the molar mass of sucrose (C12H22O11) = 342.3 g/mol
Number of moles of sucrose = mass / molar mass
Number of moles of sucrose = 7.5 g / 342.3 g/mol
Number of moles of sucrose = 0.02192 mol
Next, let's calculate the mass of water:
Given the density of water = 1 g/cm^3
Given the volume of water = 150 cm^3
Mass of water = density * volume
Mass of water = 1 g/cm^3 * 150 cm^3
Mass of water = 150 g
Now, let's 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 grams to kg)
Mass of solvent = 150 g / 1000
Mass of solvent = 0.15 kg
Molality = 0.02192 mol / 0.15 kg
Molality = 0.1461 mol/kg
Finally, let's calculate the change in freezing point temperature (ΔTf):
ΔTf = Kf * m
ΔTf = 1.86 K kg mol-1 * 0.1461 mol/kg
ΔTf = 0.2709 K
Therefore, the estimate of the freezing point of 150 cm^3 of water sweetened with 7.5 g of sucrose is approximately 0.2709 K below the normal freezing point of pure water.
To estimate the freezing point of the sweetened water, you need to use the freezing point depression equation, which states that the change in freezing point is directly proportional to the molal concentration of the solute (sucrose in this case).
First, let's calculate the molality (molal concentration) of the sucrose solution:
1. Calculate the molar mass of sucrose (C12H22O11):
- The atomic mass of carbon (C) is 12.01 g/mol.
- The atomic mass of hydrogen (H) is 1.01 g/mol.
- The atomic mass of oxygen (O) is 16.00 g/mol.
Therefore, the molar mass of sucrose is:
(12 * 12.01) + (22 * 1.01) + (11 * 16.00) = 342.29 g/mol
2. Calculate the number of moles of sucrose:
The mass of sucrose given is 7.5 g.
The number of moles can be calculated using the formula:
moles = mass / molar mass
Therefore, the number of moles of sucrose is:
7.5 g / 342.29 g/mol = 0.0219 mol
3. Calculate the molality:
Molality (m) is defined as the number of moles of solute per kilogram of solvent.
In this case, the solvent is water, and its mass is equal to its volume since its density is 1 g/cm³. Therefore, 150 cm³ of water weighs 150 g.
Therefore, the molality of the sucrose solution is:
moles of solute / mass of solvent (in kg)
molality = 0.0219 mol / 0.150 kg = 0.146 M/kg
Now that we have the molality, we can use the freezing point depression equation to estimate the change in freezing point.
ΔT = Kf * m
Where:
ΔT = change in freezing point
Kf = freezing point depression constant for water (1.86 K kg mol⁻¹)
m = molality of the solution (0.146 M/kg)
Now, substitute the values into the equation:
ΔT = 1.86 K kg mol⁻¹ * 0.146 M/kg
The unit for molality (M/kg) cancels out with the unit for the Kf constant (K kg mol⁻¹), leaving us with ΔT in Kelvin.
Finally, calculate the change in freezing point:
ΔT = 0.272 K
To obtain the estimated freezing point of the sweetened water, subtract the calculated change in freezing point from the freezing point of pure water, which is 0°C or 273.15 K.
Freezing point = 273.15 K - 0.272 K = 272.878 K
Therefore, the estimated freezing point of 150 cm³ of water sweetened with 7.5 g of sucrose is approximately 272.878 K.