Freezing point depression calculations

[physicsman2]

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.

 

[Borek]

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

 

[physicsman2]

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

 

[Borek]

I got slightly below 2k as well.

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[physicsman2]

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