100 grams of KNO3 is dissolved in one liter of water. Calculate the freezing point depression and boiling point elevation.
To calculate the freezing point depression and boiling point elevation, we need to use the properties of the solute (KNO3) and the formula for these calculations.
The freezing point depression is calculated using the formula:
∆Tf = Kf * m * i
The boiling point elevation is calculated using the formula:
∆Tb = Kb * m * i
Where:
∆Tf = freezing point depression
∆Tb = boiling point elevation
Kf = cryoscopic constant for the solvent
Kb = ebullioscopic constant for the solvent
m = molality (moles of solute per kg of solvent)
i = Van't Hoff factor (number of particles into which the solute dissociates)
For KNO3, the Van't Hoff factor (i) is 2 since it dissociates into two ions (K+ and NO3-).
First, calculate the molality (m) of the solution:
m = moles of solute / kg of solvent
Given that 100 grams of KNO3 is dissolved in 1 liter of water, we need to convert grams to moles and liters to kilograms.
The molar mass of KNO3 is:
K = 39.1 g/mol
N = 14.0 g/mol
O = 16.0 g/mol (x3 because there are three oxygen atoms)
Molar mass of KNO3 = 39.1 + 14.0 + (16.0 x 3) = 101.1 g/mol
Number of moles = mass / molar mass
Number of moles of KNO3 = 100 g / 101.1 g/mol = 0.989 moles
Since 1 liter of water is equivalent to 1 kilogram, the molality (m) is:
m = 0.989 moles / 1 kg = 0.989 m
Now, we can calculate the freezing point depression and boiling point elevation.
Freezing Point Depression:
∆Tf = Kf * m * i
The cryoscopic constant (Kf) for water is 1.86 °C/m.
∆Tf = (1.86 °C/m) * (0.989 m) * (2) = 3.684 °C
Therefore, the freezing point depression is 3.684 °C.
Boiling Point Elevation:
∆Tb = Kb * m * i
The ebullioscopic constant (Kb) for water is 0.512 °C/m.
∆Tb = (0.512 °C/m) * (0.989 m) * (2) = 1.012 °C
Therefore, the boiling point elevation is 1.012 °C.
To calculate the freezing point depression and boiling point elevation, we need to apply colligative properties. These properties depend on the number of particles (in this case, ions or molecules) present in a solution.
First, we need to find the molality (m) of KNO3 in the solution. Molality (m) is defined as the number of moles of solute per kilogram of solvent.
To find the number of moles of KNO3, we divide the mass (100 grams) by its molar mass (KNO3 = 101.1 + 14.0 + 16.0x3 = 101.1 + 42.0 = 143.1 g/mol).
Moles of KNO3 = 100 g / 143.1 g/mol = 0.698 mol
Next, we need to find the mass of water in kilograms (kg). Since the density of water is 1 g/mL or 1000 g/L, the mass of 1 L of water is 1000 g.
Mass of water = 1000 g = 1 kg
Using the moles of KNO3 and mass of water, we can calculate the molality (m):
m = moles of solute / mass of solvent (in kg)
m = 0.698 mol / 1 kg = 0.698 m
Now, we can use the molality to calculate the freezing point depression and boiling point elevation using the following equations:
For freezing point depression:
∆Tf = Kf * m
where ∆Tf is the change in freezing point, and Kf is the cryoscopic constant for water. For water, Kf is approximately 1.86 °C/m.
∆Tf = 1.86 °C/m * 0.698 m
∆Tf = 1.29628 °C
Therefore, the freezing point depression is approximately 1.30 °C.
For boiling point elevation:
∆Tb = Kb * m
where ∆Tb is the change in boiling point, and Kb is the ebullioscopic constant for water. For water, Kb is approximately 0.512 °C/m.
∆Tb = 0.512 °C/m * 0.698 m
∆Tb = 0.357376 °C
Therefore, the boiling point elevation is approximately 0.36 °C.