Freezing point depression calculations

In summary, the student is having trouble with the data and calculations section of their lab. They are unable to match the molar mass of the given solutes to their answer. The student provides the equations and their numbers, including the freezing point and molality of the solution. They also mention a possible error in the freezing point and kf values given to them by their instructor. Despite this, the student's calculations seem to be correct and they plan to explain the discrepancy in their experimental values.
  • #1
physicsman2
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0

Homework Statement


Hi, I have a problem with the data and calculations part in my lab. For some reason, I can't get a molar mass of the given solutes to match my answer. I'll provide all the equations and my numbers. Please I really need help with this

Homework Equations


deltaT = m(kf)
moles of solute=(molality)(kg of solvent used)
molar mass = mass of solute/moles of solute

The Attempt at a Solution


My freezing point for my cyclohexanol solvent was 11.0 degrees C and the freezing point I got for my solution was 9.4 degrees C(It could've been less, but my instructor told me to stop there). I found delta T to be 1.6 by subtracting the two temperatures. Then, I found the molality of the solution by dividing the change in temperature by kf, which was give to be 39.4 degrees C/m. I got .041 and went on. Next, they asked to find the moles of solute which was the molality I found to be .041 times the kg of solvent used. In grams, the solvent weighed 9.2 g, which is .0092 kg. I multiplied those two numbers and got .00038 moles. Then, they asked to find the molar mass of the solute by dividing the mass of the solute in grams, .75 g, by the moles of solute, where i got about 2000g. The only choices were 214.4g, 242.5g, and 270.5g. I can't see what I did wrong.
 
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  • #2
Melting point for cyclohexanol is not 11 but 25.4 deg C and kf of 39.4 seems to be quite high - are you sure about these numbers and solvent identity?

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methods
 
  • #3
i knew the freezing point for cyclohexanol was a lot higher, but that's what i got in my experiment. As for kf, they gave us that value. I'm not sure if I have to convert it or something, that's why i gave the units. My masses were all in grams but i converted the mass of my solvent to find the moles of the solute. My instructor said it's okay if our answers are off as long as we got a correct answer with the data we had. Can you check that I at least did it right? Thanks for responding
 
  • #4
I got slightly below 2k as well.

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  • #5
Thanks so much, i'll just have to explain the faulty experimental values
 

FAQ: Freezing point depression calculations

1. What is freezing point depression and why is it important?

Freezing point depression is a phenomenon that occurs when the freezing point of a solution is lower than that of the pure solvent. It is important because it allows us to control the freezing point of a solution, which has many practical applications in various industries such as food preservation and antifreeze solutions.

2. How is freezing point depression calculated?

Freezing point depression is calculated using the equation ∆T = Kf * m, where ∆T is the change in freezing point, Kf is the molal freezing point depression constant for the solvent, and m is the molality of the solution. This formula takes into account the colligative properties of the solution, which are based on the number of particles present in the solution rather than their individual identities.

3. What factors affect freezing point depression?

The two main factors that affect freezing point depression are the identity and concentration of the solute. Different solutes have different effects on the freezing point of a solution, and the greater the concentration of the solute, the greater the freezing point depression. Other factors that may also play a role include pressure, impurities in the solvent, and the size of the solute particles.

4. Can freezing point depression be used to determine the molar mass of a solute?

Yes, freezing point depression can be used to determine the molar mass of a solute by rearranging the freezing point depression equation to solve for the molar mass of the solute (m = ∆T / Kf * molality). This method is commonly used in chemistry labs to determine the molar mass of an unknown solute in a solution.

5. What are some real-world applications of freezing point depression calculations?

Freezing point depression calculations have many practical applications. For example, they are used in the production of ice cream to create a smoother texture, in the production of antifreeze solutions for cars, and in the preservation of food by slowing down the growth of bacteria. Freezing point depression is also important in the production of pharmaceuticals, where precise control of the freezing point of solutions is crucial for the stability and efficacy of the drugs.

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