Which of the following solutes, dissolved in 1000g of water, would provide a solution would the lowest freezing point?
.030 mol of BaCl2,
CO(NH2)2,
CaSO4,
CH3COOH,
or .030 mol of NH4NO3
Again, the steps needed to determine for the freezing point.
The solute with the lowest freezing point would be .030 mol of BaCl2. This is because BaCl2 has the highest freezing point depression of the five solutes, meaning that it will lower the freezing point of the solution the most. To calculate the freezing point of the solution, you would need to use the equation: Tf = Kf * m, where Tf is the freezing point of the solution, Kf is the freezing point depression constant for water, and m is the molality of the solution.
To determine which solute would provide a solution with the lowest freezing point, you need to calculate the molality (moles of solute per kilogram of solvent) for each solute. The solute with the highest molality will have the lowest freezing point.
Step 1: Calculate the molality for each solute.
For BaCl2:
Molar mass of BaCl2 = 137.33 g/mol
Moles of BaCl2 = 0.030 mol
Mass of water = 1000 g
Molality of BaCl2 = (0.030 mol) / (1000 g / 1000 g) = 0.030 mol/kg
For CO(NH2)2:
Molar mass of CO(NH2)2 = 60.08 g/mol
Moles of CO(NH2)2 = 1 mol
Mass of water = 1000 g
Molality of CO(NH2)2 = (1 mol) / (1000 g / 1000 g) = 1 mol/kg
For CaSO4:
Molar mass of CaSO4 = 136.14 g/mol
Moles of CaSO4 = 1 mol
Mass of water = 1000 g
Molality of CaSO4 = (1 mol) / (1000 g / 1000 g) = 1 mol/kg
For CH3COOH:
Molar mass of CH3COOH = 60.05 g/mol
Moles of CH3COOH = 1 mol
Mass of water = 1000 g
Molality of CH3COOH = (1 mol) / (1000 g / 1000 g) = 1 mol/kg
For NH4NO3:
Molar mass of NH4NO3 = 80.04 g/mol
Moles of NH4NO3 = 0.030 mol
Mass of water = 1000 g
Molality of NH4NO3 = (0.030 mol) / (1000 g / 1000 g) = 0.030 mol/kg
Step 2: Compare the molality values.
The molality values are:
BaCl2: 0.030 mol/kg
CO(NH2)2: 1 mol/kg
CaSO4: 1 mol/kg
CH3COOH: 1 mol/kg
NH4NO3: 0.030 mol/kg
The solute with the highest molality is CO(NH2)2, which means it will provide a solution with the lowest freezing point.
To determine which solute would result in the lowest freezing point of a solution, you need to understand the concept of colligative properties. One colligative property is the freezing point depression, which is the difference between the freezing point of the pure solvent and the freezing point of the solution.
The formula to calculate the freezing point depression is:
ΔTf = i * Kf * m
Where:
- ΔTf is the freezing point depression
- i is the van't Hoff factor (the number of particles into which the solute dissociates in solution)
- Kf is the cryoscopic constant (a characteristic property of the solvent)
- m is the molality of the solute (moles of solute per kilogram of solvent)
Since the molar mass is not provided for the solutes, let's calculate the molality of each solute:
1. BaCl2:
Given: 0.030 mol of BaCl2
Molar mass of BaCl2: 137.33 g/mol
Molality (m) = moles of solute / mass of solvent (in kg)
= 0.030 mol / 1 kg
= 0.030 mol/kg
2. CO(NH2)2:
Molar mass of CO(NH2)2: 60.08 g/mol
Molality (m) = moles of solute / mass of solvent (in kg)
= 0.030 mol / 1 kg
= 0.030 mol/kg
3. CaSO4:
Molar mass of CaSO4: 136.14 g/mol
Molality (m) = moles of solute / mass of solvent (in kg)
= 0.030 mol / 1 kg
= 0.030 mol/kg
4. CH3COOH:
Molar mass of CH3COOH: 60.05 g/mol
Molality (m) = moles of solute / mass of solvent (in kg)
= 0.030 mol / 1 kg
= 0.030 mol/kg
5. NH4NO3:
Given: 0.030 mol of NH4NO3
Molar mass of NH4NO3: 80.04 g/mol
Molality (m) = moles of solute / mass of solvent (in kg)
= 0.030 mol / 1 kg
= 0.030 mol/kg
Now that we have the molality values for each solute, we can compare the freezing point depressions by using the formula:
ΔTf = i * Kf * m
For each solute, you would need to look up the van't Hoff factor (i) and the cryoscopic constant (Kf) for water.
The solute with the highest freezing point depression (ΔTf) will result in the lowest freezing point.