calculate the freezing point of an aqueous solution containing 10.50 g of MgBr2 in 200 g of water (MgBr2=184 g, Kf for water =1.80 K kg mol-1)
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To calculate the freezing point of an aqueous solution, we need to use the equation:
ΔTf = Kf·m
Where:
ΔTf = change in freezing point
Kf = freezing point depression constant for the solvent (water in this case)
m = molality of the solute (MgBr2) in the solution
First, let's calculate the number of moles of MgBr2 in the solution:
Number of moles of MgBr2 = mass of MgBr2 / molar mass of MgBr2
Given:
mass of MgBr2 = 10.50 g
molar mass of MgBr2 = 184 g/mol
Number of moles of MgBr2 = 10.50 g / 184 g/mol
Number of moles of MgBr2 = 0.057 moles
Next, let's calculate the molality (m) of the solution:
Molality (m) = moles of solute / mass of solvent (in kg)
Given:
mass of water = 200 g = 0.2 kg
Molality (m) = 0.057 moles / 0.2 kg
Molality (m) = 0.285 mol/kg
Now, we can calculate the freezing point depression (ΔTf):
ΔTf = Kf·m
Given:
Kf for water = 1.80 K kg mol^-1
ΔTf = 1.80 K kg mol^-1 · 0.285 mol/kg
ΔTf = 0.513 K
Finally, we can calculate the freezing point of the solution:
Freezing point = Freezing point of pure solvent - ΔTf
The freezing point of pure water is 0°C or 273.15 K.
Freezing point = 273.15 K - 0.513 K
Freezing point = 272.637 K
Therefore, the freezing point of the solution containing 10.50 g of MgBr2 in 200 g of water is 272.637 K.
To calculate the freezing point of the aqueous solution, we need to use the formula for freezing point depression:
ΔT = Kf * m
where:
ΔT is the freezing point depression
Kf is the freezing point depression constant for water, which is given as 1.80 K kg mol-1
m is the molality of the solution, which is calculated by dividing the moles of solute by the mass of the solvent in kilograms
First, let's calculate the moles of MgBr2:
moles of MgBr2 = mass of MgBr2 / molar mass of MgBr2
mass of MgBr2 = 10.50 g
molar mass of MgBr2 = 2 * atomic mass of Mg + 2 * atomic mass of Br = 2 * 24.31 g/mol + 2 * 79.90 g/mol = 184 g/mol
moles of MgBr2 = 10.50 g / 184 g/mol
Next, let's calculate the mass of water in kilograms:
mass of water = 200 g = 200 g / 1000 = 0.200 kg
Now, we can calculate the molality of the solution:
molality = moles of MgBr2 / mass of water in kg
Next, we can calculate the freezing point depression:
ΔT = Kf * m
Finally, we can calculate the freezing point of the solution by subtracting the freezing point depression from the freezing point of pure water (0°C or 273.15 K).
Freezing point of solution = Freezing point of water - ΔT
Now, let's plug in the values and calculate the freezing point of the solution:
moles of MgBr2 = 10.50 g / 184 g/mol = 0.057 moles
mass of water = 0.200 kg
molality = 0.057 moles / 0.200 kg = 0.285 mol/kg
ΔT = 1.80 K kg mol-1 * 0.285 mol/kg = 0.513 K
Freezing point of solution = 273.15 K - 0.513 K = 272.637 K
Therefore, the freezing point of the aqueous solution containing 10.50 g of MgBr2 in 200 g of water is approximately 272.637 K.