Determine the freezing points of each 0.20m aqueous solution.
a)K2SO4
b)CsNO3
c)Al(NO3)3
a) Well, if K2SO4 were a superhero, its freezing point would probably be "super low". That's because K2SO4 is an ionic compound that breaks up into potassium cations (K+) and sulfate anions (SO4^2-), and these don't make very good snuggle buddies for water molecules. So, I would say the freezing point of a 0.20m K2SO4 solution would be below freezing, somewhere in the "icy" realm.
b) CsNO3 is like the cool kid at the freezing point party. CsNO3 is also an ionic compound that dissolves in water, forming cesium cations (Cs+) and nitrate anions (NO3^-). These ions mingle pretty well with the water, making the solution less likely to freeze easily. So, the freezing point of a 0.20m CsNO3 solution would be higher than that of pure water. CsNO3 is all about keeping things chill.
c) Ah, Al(NO3)3, should we call it the "aluminum freeze stopper"? Well, that might be a bit of an exaggeration, but Al(NO3)3 is a salt that does lower the freezing point of water when dissolved. When it dissolves, it breaks up into aluminum cations (Al^3+) and nitrate anions (NO3^-), which disrupt the water molecules' ability to pack together nicely and form a solid. So, the freezing point of a 0.20m Al(NO3)3 solution would be lower than that of pure water. Aluminum definitely has the power to keep water liquid at colder temperatures.
To determine the freezing points of the given aqueous solutions, we need to calculate the expected change in freezing point, which is determined by the number of particles in the solution.
The formula to calculate the change in freezing point is:
∆Tf = i * Kf * m
Where:
∆Tf is the change in freezing point
i is the van't Hoff factor (number of particles)
Kf is the cryoscopic constant of the solvent
m is the molality of the solution
For all the solutions, we'll assume the cryoscopic constant (Kf) for water is -1.86 °C/m.
a) K2SO4:
K2SO4 dissociates into three particles: 2K+ and SO4^2- ions. Therefore, the van't Hoff factor (i) for K2SO4 is 3.
Given the molality (m) is 0.20 m, we can calculate the change in freezing point (∆Tf) as follows:
∆Tf = 3 * (-1.86 °C/m) * (0.20 m)
∆Tf = -1.116 °C
The freezing point of the 0.20m K2SO4 solution is then the freezing point of pure water (0 °C) minus the change in freezing point:
Freezing point = 0 °C - 1.116 °C
Freezing point = -1.116 °C
Therefore, the freezing point of the 0.20m K2SO4 solution is approximately -1.116 °C.
b) CsNO3:
CsNO3 dissociates into two ions: Cs+ and NO3-. Therefore, the van't Hoff factor (i) for CsNO3 is 2.
Using the same cryoscopic constant (Kf) and molality (m) as above, we can calculate the change in freezing point (∆Tf) as follows:
∆Tf = 2 * (-1.86 °C/m) * (0.20 m)
∆Tf = -0.744 °C
The freezing point of the 0.20m CsNO3 solution is then:
Freezing point = 0 °C - 0.744 °C
Freezing point = -0.744 °C
Therefore, the freezing point of the 0.20m CsNO3 solution is approximately -0.744 °C.
c) Al(NO3)3:
Al(NO3)3 dissociates into four ions: Al^3+ and three NO3- ions. Therefore, the van't Hoff factor (i) for Al(NO3)3 is 4.
Using the same cryoscopic constant (Kf) and molality (m) as above, we can calculate the change in freezing point (∆Tf) as follows:
∆Tf = 4 * (-1.86 °C/m) * (0.20 m)
∆Tf = -1.488 °C
The freezing point of the 0.20m Al(NO3)3 solution is then:
Freezing point = 0 °C - 1.488 °C
Freezing point = -1.488 °C
Therefore, the freezing point of the 0.20m Al(NO3)3 solution is approximately -1.488 °C.
To determine the freezing points of the aqueous solutions, you need to use the concept of freezing point depression. Freezing point depression occurs when a solute is added to a solvent, causing the freezing point of the solution to be lower than that of the pure solvent.
The formula to calculate freezing point depression is:
ΔTf = Kf * m
where:
ΔTf = freezing point depression (in ˚C)
Kf = cryoscopic constant or molal freezing point depression constant (unique to each solvent, measured in ˚C/m)
m = molality of the solution (moles of solute per kilogram of solvent)
To get the freezing points of the 0.20m aqueous solutions of the given solutes, you will need to know the cryoscopic constants for water (the solvent) at that concentration range. The cryoscopic constant for water at 0.20 m is approximately 1.86 ˚C/molal.
Now, let's calculate the freezing point depressions for each of the solutions:
a) K2SO4:
The compound K2SO4 dissociates into three ions in water: 2K+ ions and 1 SO4^2- ion. This means that the effective concentration of particles in the solution is 3 times the concentration of K2SO4.
ΔTf = Kf * m
ΔTf = 1.86 ˚C/molal * 0.20 m * 3
ΔTf = 0.1116 ˚C
The freezing point of the 0.20m aqueous solution of K2SO4 would be 0 ˚C - 0.1116 ˚C = -0.1116 ˚C.
b) CsNO3:
The compound CsNO3 dissociates into two ions in water: 1 Cs+ ion and 1 NO3- ion. This means that the effective concentration of particles in the solution is 2 times the concentration of CsNO3.
ΔTf = Kf * m
ΔTf = 1.86 ˚C/molal * 0.20 m * 2
ΔTf = 0.744 ˚C
The freezing point of the 0.20m aqueous solution of CsNO3 would be 0 ˚C - 0.744 ˚C = -0.744 ˚C.
c) Al(NO3)3:
The compound Al(NO3)3 dissociates into four ions in water: 1 Al^3+ ion and 3 NO3- ions. This means that the effective concentration of particles in the solution is 4 times the concentration of Al(NO3)3.
ΔTf = Kf * m
ΔTf = 1.86 ˚C/molal * 0.20 m * 4
ΔTf = 1.488 ˚C
The freezing point of the 0.20m aqueous solution of Al(NO3)3 would be 0 ˚C - 1.488 ˚C = -1.488 ˚C.
Therefore, the freezing point of the 0.20m aqueous solution of:
a) K2SO4 is -0.1116 ˚C
b) CsNO3 is -0.744 ˚C
c) Al(NO3)3 is -1.488 ˚C
Same process as your last post.
Just use delta T = i*Kf*m