Arrange the following aqueous solutions in order of increasing boiling point, explain your answer: 0.120 m glucose (C6H12O6), 0.05 m LiBr, 0.05 m Zn(NO3)2. Using data the boiling point constants given in the test book, calculate the boiling points of each of the above aqueous solutions to see if your predictions were correct.

delta T = i*Kb*m

i = 1 for glucose
i = 2 for LiBr
i = 3 for Zn(NO3)2

Kb is constant so delta T is a function of i*m. Those will arrange for you.
You can plug the numbers into the formula and check for yourself. Post your work if you get stuck.

LiBr<Glucose<Zinc nitrate

To determine the order of increasing boiling point for the given aqueous solutions, we need to consider the colligative properties of the solutes. The boiling point elevation is a colligative property that depends on the number of solute particles present in the solution.

The boiling point elevation (ΔTb) for an aqueous solution can be calculated using the formula:

ΔTb = Kb * m

where ΔTb is the boiling point elevation, Kb is the boiling point constant (different for each solute), and m is the molality of the solution.

Now, let's analyze each solution and calculate the boiling points:

1. 0.120 m glucose (C6H12O6):
Since glucose is a nonelectrolyte (does not dissociate into ions), it will not increase the boiling point significantly. Therefore, this solution is expected to have the lowest boiling point.

2. 0.05 m LiBr:
LiBr is an ionic compound that dissociates into Li+ and Br- ions in water. The presence of ions increases the boiling point of the solution more significantly than non-electrolytes.

3. 0.05 m Zn(NO3)2:
Zn(NO3)2 is also an ionic compound that dissociates into Zn2+ and 2 NO3- ions. As with LiBr, the presence of more ions will lead to a larger boiling point increase.

Based on this analysis, the order of increasing boiling point for the given solutions is:
0.120 m glucose (C6H12O6) < 0.05 m LiBr < 0.05 m Zn(NO3)2

To calculate the boiling points, we can use the boiling point elevation formula mentioned earlier. We need the boiling point constant (Kb) for each solute. However, you mentioned that it is given in the test book, so please refer to your specific book or reference material to obtain the Kb values.

Once you have the Kb values, you can use the formula ΔTb = Kb * m to calculate the boiling point elevation for each solution. Then, add the boiling point elevation to the normal boiling point of water (100 °C or 212 °F) to obtain the boiling point of each solution.

To determine the order of increasing boiling point for the given aqueous solutions, we need to consider the concept of boiling point elevation, which is a colligative property of the solutions. According to Raoult's Law, the boiling point elevation is directly proportional to the molality (m) of the solute in the solution.

To calculate the boiling points, we need to know the boiling point constant (Kb) for water, which is usually provided in the textbook or given in the question. With the Kb, we can use the following formula to calculate the boiling point elevation (∆Tb):

∆Tb = Kb * m,

where m is the molality of the solute in the solution.

Let's calculate the boiling points of each solution and then determine the order of increasing boiling point:

1. 0.120 m glucose (C6H12O6):
Using the given molality (m = 0.120), we can calculate ∆Tb using the boiling point constant for water (Kb):
∆Tb = Kb * m.

2. 0.05 m LiBr:
Again, using the given molality (m = 0.05), we can calculate ∆Tb using Kb.

3. 0.05 m Zn(NO3)2:
Similarly, using the given molality (m = 0.05), we can calculate ∆Tb using Kb.

After calculating the boiling points of the solutions, we can compare them and arrange the solutions in order of increasing boiling point.

Please provide the boiling point constants (Kb) mentioned in the textbook, and I can help you calculate the boiling points and determine the order.