Arrange the following aqueous solutions in order of decreasing freezing points (lowest to highest temperature): 0.10 m Na3PO4, 0.35 m NaCl, 0.20 m MgCl, 0.15 m C6H12O6 and 0.15 m CH3COOH.
delta T = i*Kf*m
i is the van't Hoff factor which is 4 for Na3PO4, 2 for NaCl, 3 for MgCl2(you made a typo here), 1 for C6H12O6 and 1 for CH3COOH.
Example for Na3PO4.
delta T = 4*1.86*0.1 = ?
For NaCl:
delta T = 2*1.86*0.35 = ?
Then freezing point = 0 - delta T = ?
So which has the lower freezing point.
Calculate all of them then arrange in order.
C6H12O6>CH3COOH>Na3PO4>MgCl2>NaCl
To compare the freezing points of aqueous solutions, we need to look at the concentration of solute particles in each solution.
First, let's count the number of ions or particles that each solute will dissociate into in water:
1. Na3PO4 dissociates into 4 ions: 3 Na+ ions and 1 PO4^3- ion.
2. NaCl dissociates into 2 ions: 1 Na+ ion and 1 Cl- ion.
3. MgCl2 dissociates into 3 ions: 1 Mg^2+ ion and 2 Cl- ions.
4. C6H12O6 (glucose) does not dissociate into ions and remains as individual molecules.
5. CH3COOH (acetic acid) partially dissociates into 2 ions: 1 H+ ion and 1 CH3COO- ion.
Now, let's arrange them in order of decreasing freezing points:
1. C6H12O6: Since it does not dissociate into ions, it will have the least number of particles in solution, resulting in the highest freezing point.
2. CH3COOH: Although it partially dissociates, it still has 2 ions, which is more than glucose. Therefore, it will have a lower freezing point than glucose.
3. NaCl: It dissociates fully into 2 ions, so it will have a lower freezing point than acetic acid.
4. MgCl2: It dissociates into 3 ions, which is more than sodium chloride. Thus, it will have a lower freezing point than sodium chloride.
5. Na3PO4: It dissociates into 4 ions, which is the highest number among all the solutes. Therefore, it will have the lowest freezing point.
Based on this analysis, the solutions can be arranged in order of decreasing freezing points as follows:
0.10 m Na3PO4 > 0.15 m MgCl2 > 0.35 m NaCl > 0.15 m CH3COOH > 0.15 m C6H12O6.
To determine the order of decreasing freezing points for the given aqueous solutions, you need to consider the concept of freezing point depression.
Freezing point depression is a colligative property that depends on the number of solute particles present in a solution, rather than the identity of the solute. The greater the concentration of solute particles, the lower the freezing point of the solution compared to the pure solvent.
In this case, we have five aqueous solutions: 0.10 m Na3PO4, 0.35 m NaCl, 0.20 m MgCl, 0.15 m C6H12O6, and 0.15 m CH3COOH.
First, let's count the number of solute particles for each compound by determining the number of dissociated ions (if applicable) or individual solute particles. This will give us an idea of the resulting freezing point depression.
1. Na3PO4:
Na3PO4 dissociates into three Na+ ions and one PO4(3-) ion, giving a total of four solute particles.
2. NaCl:
NaCl dissociates into one Na+ ion and one Cl- ion, giving a total of two solute particles.
3. MgCl:
MgCl does not dissociate, so it remains as one solute particle.
4. C6H12O6 (glucose):
C6H12O6 remains as one molecule, so it also remains as one solute particle.
5. CH3COOH (acetic acid):
CH3COOH can partially dissociate into H+ ions and CH3COO- ions. However, since the concentration is relatively low (0.15 m), we can consider that it remains mainly as one molecule and therefore one solute particle.
Now let's arrange the solutions in order of decreasing freezing points based on the number of solute particles:
1. Na3PO4 (four solute particles)
2. NaCl (two solute particles)
3. MgCl (one solute particle)
4. C6H12O6 (one solute particle)
5. CH3COOH (one solute particle)
So, the correct order of decreasing freezing points (lowest to highest temperature) for the given solutions is:
Na3PO4 < NaCl < MgCl < C6H12O6 = CH3COOH