Assuming 100% dissociation, calculate the freezing point and boiling point of 0.700 mol of AgNO3 in 1.00 kg of water.
Well, let me tell you, AgNO3 is not known for its sense of humor. But let's be serious for a minute and calculate these values for you!
To calculate the freezing point depression, we will need to use the equation:
ΔTf = Kf * m
Where:
ΔTf is the freezing point depression,
Kf is the molal freezing point depression constant for water (-1.86°C/m),
and m is the molality of the solution.
First things first, let's calculate the molality (m):
m = moles of solute / mass of solvent (in kg)
Given that you have 0.700 mol of AgNO3 and 1.00 kg of water (which is 1000 g), we can do the math:
m = 0.700 mol / 1.000 kg = 0.700 m
Now, using the formula above, let's calculate the freezing point depression (ΔTf):
ΔTf = (-1.86°C/m) * (0.700 m) = -1.302 °C
Now for the fun part, let's calculate the new freezing point:
Freezing point = normal freezing point of water - ΔTf
The normal freezing point of water is 0°C, so:
Freezing point = 0°C - (-1.302°C) = 1.302°C
As for the boiling point elevation, we can use a similar equation:
ΔTb = Kb * m
Where:
ΔTb is the boiling point elevation,
Kb is the molal boiling point elevation constant for water (0.52°C/m),
and m is the molality of the solution (which we already calculated as 0.700).
Using the formula, we can calculate the boiling point elevation (ΔTb):
ΔTb = (0.52°C/m) * (0.700 m) = 0.364 °C
And finally, let's calculate the new boiling point:
Boiling point = normal boiling point of water + ΔTb
The normal boiling point of water is 100°C, so:
Boiling point = 100°C + 0.364°C = 100.364°C
So, assuming 100% dissociation, the freezing point of the solution would be 1.302°C and the boiling point would be 100.364°C. Enjoy melting your ice and boiling water with AgNO3, just be careful not to clown around too much!
To calculate the freezing and boiling points of a solution, we need to use the formula for freezing point depression and boiling point elevation.
Freezing Point Depression:
ΔTf = Kf * molality
Boiling Point Elevation:
ΔTb= Kb * molality
Where:
ΔTf = change in freezing point
ΔTb = change in boiling point
Kf = freezing point depression constant (for water, it is 1.86 °C/m)
Kb = boiling point elevation constant (for water, it is 0.512 °C/m)
molality = moles of solute / mass of solvent (in kg)
Given:
mole of AgNO3 = 0.700 mol
mass of water = 1.00 kg
Step 1: Calculate the molality
molality = moles of solute / mass of solvent (in kg)
molality = 0.700 mol / 1.00 kg
molality = 0.700 mol/kg
Step 2: Calculate the change in freezing point (ΔTf)
ΔTf = Kf * molality
ΔTf = 1.86 °C/m * 0.700 mol/kg
ΔTf = 1.302 °C
Step 3: Calculate the change in boiling point (ΔTb)
ΔTb = Kb * molality
ΔTb = 0.512 °C/m * 0.700 mol/kg
ΔTb = 0.358 °C
Step 4: Calculate the freezing point of the solution
Freezing point of pure water = 0 °C
Freezing point of solution = Freezing point of pure water - ΔTf
Freezing point of solution = 0 °C - 1.302 °C
Freezing point of solution = -1.302 °C
Step 5: Calculate the boiling point of the solution
Boiling point of pure water = 100 °C
Boiling point of solution = Boiling point of pure water + ΔTb
Boiling point of solution = 100 °C + 0.358 °C
Boiling point of solution = 100.358 °C
Therefore, assuming 100% dissociation, the freezing point of the solution is -1.302 °C and the boiling point of the solution is 100.358 °C.
To calculate the freezing and boiling points of a solution, we need to consider the molality of the solute and the properties of the solvent. In this case, the solute is 0.700 mol of AgNO3, and the solvent is 1.00 kg of water.
First, we need to find the molality of the solution. Molality (m) is defined as the moles of solute per kilogram of solvent.
Molality (m) = moles of solute / mass of solvent (in kg)
Given that the mass of the solvent is 1.00 kg and the moles of solute is 0.700 mol, we can calculate the molality.
m = 0.700 mol / 1.00 kg = 0.700 m
Next, we need to use the molality of the solution to calculate the freezing and boiling point changes using the formula:
ΔTf = -Kf * m
ΔTb = Kb * m
Where:
ΔTf = freezing point depression
ΔTb = boiling point elevation
Kf = freezing point depression constant (for water, Kf = 1.86 °C/m)
Kb = boiling point elevation constant (for water, Kb = 0.51 °C/m)
For the freezing point depression:
ΔTf = -1.86 °C/m * 0.700 m = -1.302 °C
For the boiling point elevation:
ΔTb = 0.51 °C/m * 0.700 m = 0.357 °C
Finally, we can calculate the freezing and boiling points of the solution by adding/subtracting the freezing and boiling point changes from the normal freezing and boiling points of water.
For water, the normal freezing point is 0 °C and the normal boiling point is 100 °C.
Freezing point of the solution:
Freezing point = 0 °C - (-1.302 °C) = 1.302 °C
Boiling point of the solution:
Boiling point = 100 °C + 0.357 °C = 100.357 °C
Therefore, assuming 100% dissociation, the freezing point of the solution is 1.302 °C and the boiling point is 100.357 °C.
delta T = i*Kf*m
Substitute into the equation and solve for delta T, then subtract from 0c to find the new freezing point. i for AgNO3 is 2.
same thing for delta T = i*Kb*m but ADD delta T to normal boiling point of H2O to find the new boiling point.
m = mols/kg solvent = ?