physicsman2
Nov15-09, 12:09 AM
1. The problem statement, all variables and given/known data
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
2. Relevant equations
deltaT = m(kf)
moles of solute=(molality)(kg of solvent used)
molar mass = mass of solute/moles of solute
3. 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.
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
2. Relevant equations
deltaT = m(kf)
moles of solute=(molality)(kg of solvent used)
molar mass = mass of solute/moles of solute
3. 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.