# Chemistry

3.92g of a nondissociating compound are dissolved in 400.g of liquid benzene. The freezing point of the solution is 5.19 degrees celsius. Calculate the molar mass of the compound. (The freezing point of the pure benzene is 5.50 degrees celsius and its molal freezing point depression is 5.12 degrees celsius x kg/mole. (Answer= 1.6 x 10^2 g/mol)

Delta T = Kf mass of solute
--------------------
mole mass solute x kg solvent

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1. I find it better to work in steps.
delta T = Kf*molality
Solve for molality

molality = moles solute/kg solvent
Solve for moles solute

moles solute = grams/molar mass
solve for molar mass.

If you still have trouble, post your work on these three steps and someone will look at it for errors.

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2. Delta T = Kf x molality
5.19 C = 5.12 C kg/mol x molality
molality = 1.01 moles/kg

Molality = moles solute/kg solvent
1.01 = moles solute/0.4
moles solute= .404

Moles solute=grams/molar mass
.404=3.92g/molar mass
molar mass= .103

I think I am still doing this wrong.

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3. Your problem is that you are not substituting delta T. Instead, you are substituting the freezing point of the solution.
delta T = (difference between freezing point of the pure solvent and freezing point of the solution). In this case that is 5.50-5.19 = ??
I took a quick look at the other steps. I think those are right if you correct step #1 then go through the rest of the problem.

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4. Oh snap. Thanks so much :D
I knew I was doing something wrong.

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5. how do you figure out the moled without its molecular mass

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6. moles*

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7. you forgot to subtract the final temp with the initial temp. 5.19-5.50= 0.31
then follow the calculations as you did and the molar mass will end up being 163.33

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