Arrange the following aqueous solutions in order of increasing freezing points (lowest to highest temperature): 0.10 m glucose, 0.10 m BaCl2, 0.20 m NaCl, and 0.20 m Na2SO4.
delta T = i*Kf*molality
i = 1 for glucose
i = 3 for BaCl2
i = 2 for NaCl
i = 3 for Na2SO4
You CAN calculate each with the data given and arrange them from the answers OR you can look at i*m and see which is the largest. The largest value of i*m will have the lowest freezing point (the largest delta T) and the smallest value of i*m will have the highest freezing point (smallest delta T).
What is the boiling point of a solution made by mixing 115 g NaCl in 1 kg of water? (Kb for water is 0.512 C/m)
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.
1- Calculate the molality of the commercially available reagent HCI (36.465 amu, d=1.1 g/ml, 37.25 weight, 12.1 M)?
Sure, let's break the ice and arrange these solutions in order of increasing freezing points.
First up, we have 0.10 m glucose. Well, glucose may be sweet, but it doesn't have much "freezing point lowering" power. So, it's safe to say that it'll have the highest freezing point of the bunch.
Next, we have 0.10 m BaCl2. BaCl2 might sound like it has a bar of lower freezing point, but it's not the strongest "freezer melter" either. So, we'll place it in the second spot.
Now, let's move on to 0.20 m NaCl. Ah, good ol' table salt. Known for its ability to make icy sidewalks more manageable, NaCl has a significant freezing point lowering effect. It takes the third spot on our chilly ranking.
Last, but not least, we have 0.20 m Na2SO4. Sodium sulfate might seem like a tough contender, but its freezing point-lowering superpowers are quite mighty, especially when compared to glucose and BaCl2. Therefore, it secures the lowest freezing point in the lineup.
So, here's the order of increasing freezing points: 0.10 m glucose, 0.10 m BaCl2, 0.20 m NaCl, and 0.20 m Na2SO4. Stay cool!
To determine the order of increasing freezing points for the given aqueous solutions, we need to consider the concept of freezing point depression. Freezing point depression occurs when a solute is added to a solvent, reducing the temperature at which the solution freezes compared to the pure solvent.
The formula to calculate the freezing point depression is: ΔTf = Kf * m
Where:
- ΔTf is the change in freezing point (in degrees Celsius)
- Kf is the cryoscopic constant (a characteristic of the solvent)
- m is the molality of the solution (moles of solute per kilogram of solvent)
The greater the molality (m), the greater the freezing point depression. Therefore, the largest molality will lead to the lowest freezing point.
Now let's compare the given solutions:
0.10 m glucose
0.10 m BaCl2
0.20 m NaCl
0.20 m Na2SO4
We can see that both NaCl and Na2SO4 have the same molality of 0.20 m, which means they will have the greatest freezing point depression.
Next, we have 0.10 m glucose and 0.10 m BaCl2. Comparing these two, we need to consider the type of solute.
Glucose is a non-electrolyte, which means it does not dissociate into ions in water. On the other hand, BaCl2 will dissociate into three ions in water: Ba^2+, and two Cl^- ions.
Since glucose does not dissociate, it will have a lower freezing point depression compared to BaCl2, which dissociates into more particles. Therefore, 0.10 m glucose will have a higher freezing point than 0.10 m BaCl2.
Based on the above analysis, we can arrange the solutions in order of increasing freezing points (lowest to highest temperature):
0.10 m BaCl2
0.10 m glucose
0.20 m NaCl
0.20 m Na2SO4