four flasks each contain 100 milliliters of aqueous solutions equal concentrations at 25 degress C and 1 atm. the four flasks are KCL,CH3OH, Ba(OH)2, and CH3COOH.
the question is Which solution had the lowest freezing point and explain answer
To determine which solution has the lowest freezing point, we need to consider the freezing point depression caused by each solute.
The freezing point depression is related to the molality of the solute present in the solution. Molality is calculated by dividing the moles of solute by the mass of the solvent in kilograms.
First, we need to determine the moles of each solute present in the 100 ml solution.
1. For KCl:
We need to find the moles of KCl in 100 ml of solution.
The molar mass of KCl is 74.55 g/mol.
To calculate the moles, we divide the mass of KCl by its molar mass.
Assuming the density of the KCl solution is approximately 1 g/ml, its mass would be 100g.
Moles of KCl = 100 g / 74.55 g/mol = 1.34 mol
2. For CH3OH:
The molar mass of CH3OH is 32.04 g/mol.
Moles of CH3OH = 100 g / 32.04 g/mol = 3.12 mol
3. For Ba(OH)2:
The molar mass of Ba(OH)2 is (137.33 g/mol + 2 * 16.00 g/mol + 2 * 1.01 g/mol) = 171.33 g/mol.
Moles of Ba(OH)2 = 100 g / 171.33 g/mol = 0.584 mol
4. For CH3COOH:
The molar mass of CH3COOH is (12.01 g/mol + 2 * 1.01 g/mol + 16.00 g/mol) = 60.05 g/mol.
Moles of CH3COOH = 100 g / 60.05 g/mol = 1.66 mol
Next, we need to calculate the molality of each solution by dividing the moles of solute by the mass of the solvent in kilograms.
Assuming the density of water is approximately 1 g/ml, we have 100 ml of water, which is equal to 0.1 kg.
1. Molality of KCl = 1.34 mol / 0.1 kg = 13.4 mol/kg
2. Molality of CH3OH = 3.12 mol / 0.1 kg = 31.2 mol/kg
3. Molality of Ba(OH)2 = 0.584 mol / 0.1 kg = 5.84 mol/kg
4. Molality of CH3COOH = 1.66 mol / 0.1 kg = 16.6 mol/kg
The freezing point depression of a solution is directly proportional to the molality of the solute. Therefore, the solution with the highest molality will have the lowest freezing point.
From the molalities calculated above, we can see that CH3OH has the highest molality of 31.2 mol/kg. This indicates that CH3OH has the highest freezing point depression, and subsequently the lowest freezing point.
Hence, the solution of CH3OH has the lowest freezing point among the four solutions: KCl, CH3OH, Ba(OH)2, and CH3COOH.
To determine which solution has the lowest freezing point among KCl, CH3OH, Ba(OH)2, and CH3COOH, we need to consider the concept of freezing point depression.
Freezing point depression occurs when a solute is dissolved in a solvent, reducing the freezing point of the resulting solution compared to the pure solvent. The extent of freezing point depression depends on the number of solute particles present in the solution.
To find the solution with the lowest freezing point, we need to compare the number of particles each solute will create in the solution.
1. KCl: KCl is an ionic compound, and when it dissolves in water, it dissociates into two particles: K+ and Cl-. So, one mole of KCl in solution will produce two particles.
2. CH3OH: CH3OH is a covalent compound and does not dissociate into ions when dissolved in water. Instead, it remains as individual molecules. So, one mole of CH3OH in solution will produce one particle.
3. Ba(OH)2: Ba(OH)2 is an ionic compound and will dissociate into three particles: one Ba2+ ion and two OH- ions. So, one mole of Ba(OH)2 in solution will produce three particles.
4. CH3COOH: CH3COOH is a weak acid and partially ionizes in water. It dissociates into one H+ ion and one CH3COO- ion. So, one mole of CH3COOH in solution will produce two particles.
Now, to compare the freezing point depression caused by each solute, we need to consider the Van't Hoff factor (i). The Van't Hoff factor represents the number of particles the solute dissociates into in solution.
The Van't Hoff factor for KCl is 2 since it dissociates into two particles.
The Van't Hoff factor for CH3OH is 1, as it remains as individual molecules.
The Van't Hoff factor for Ba(OH)2 is 3, as it dissociates into three particles.
The Van't Hoff factor for CH3COOH is 2, as it dissociates into two particles.
Now, the solution with the lowest freezing point will have the highest Van't Hoff factor. In this case, Ba(OH)2 has the highest Van't Hoff factor of 3, indicating that it will cause the greatest freezing point depression. Therefore, Ba(OH)2 has the lowest freezing point among the given solutions.
To summarize:
- KCl has a Van't Hoff factor of 2.
- CH3OH has a Van't Hoff factor of 1.
- Ba(OH)2 has a Van't Hoff factor of 3.
- CH3COOH has a Van't Hoff factor of 2.
Therefore, Ba(OH)2 solution will have the lowest freezing point out of the four solutions given.
delta T = i*Kb*molality.
You CAN calculate delta T for all four, then you will know quantitatively which is the lowest (actually you can't because concn is not given). BUT, you can do it the easy way. You won't know exactly what the freezing point is but you will know which is the lowest. To do it the easy, note that Kb is the same for all and the volume is the same for all, the pressure is the same for all, concentrations are the same for all. So the only thing different is the i, which is the number of particles formed when the compounds ionize. KCl forms two (K^+ and Cl^-), CH3OH doesn't ionize so it has 1, etc. The one with the most particles has the lowest freezing point because delta T will be the greatest.