[chay722]

I am asked to calculate moles of benzoic acid solute, using the molality and the mass in kg of lauric acid solvent. The molality of the lauric acid solvent is .06 mol/kg and the mass is .008 kg. I have no idea how using the molality and mass of the solvent I can solve for the solute information. Can anyone explain this?

[DUDE778]

All i can think of is to use the moles determined by muliplying the molality and 0.008 kg.
Having x number of moles can easily have us use the ratioo and get us to benzoic acid.
i hope that helps.

[chay722]

I solved for the moles of lauric acid by doing what you said but do you know what ratio I would use to solve for the moles of benzoic acid. Would I just assume they were proportional? Thanks for the help.

[enahs]

Maybe I am miss reading the question, but I am not sure what the problem is?

You said you have 0.6 mol of Benzoic Acid per 1 kg of lauric acid. So if you have 0.008 kg of the solution, how many mols of benzoic acid is in it?

[chay722]

The .06 mol/kg was the molality of the lauric acid. I need to find the moles of the benzoic acid solute.

[enahs]

What is the definition of molality? Do those mol's in the numerator represent the solvent or the solute?

[chay722]

Molality is the number of moles solute in kg solvent. I guess I did the first question incorrectly. It says calculat the molality, using the formula deltat=Kf*m. I solved for m and got .06 mol/kg. I assume this would be the molality of the lauric acid because the problem gave me the Kf value for lauric acid as 3.9 deg. Ckg/mol. Would the molality I calculated then be that of the benzoic acid?

[enahs]

You are confusing things.

The Molality is the Moles of solute in a given solvent.

The Lauric Acid is the solvent, the benzoic acid is the solute.

There will be no Freezing point depression of the pure solvent. You calculated the molality of the SOLUTION. The molality of the solution is the moles of solute per kilogram of solvent.



So if I give you 0.008kg of this solution, how many moles of benzoic acid are in it?

[chay722]

Should I set up the equation like this to determine the number of moles. .008kg(.06mol/kg)=4.8*10^-4 mol. And that would be my answer?

[chay722]

or should I divide that 4.8*10^-4 number by .001kg (the mass of the solute)?

[enahs]

or should I divide that 4.8*10^-4 number by .001kg (the mass of the solute)?

Wait? What? Where did this number (0.001kg) come from? That was not in your original question.

Call me crazy, but if you know the mass of a substance, and what that substance is (and thus its molecular formula), you know how many mols of the substance you have, no?


Maybe you should start over and type out the question completely to make sure you are being understood correctly or not missing something.  Because based on your original post, yes I would have said 0.48 milli-moles.

[chay722]

.001 kg is the mass of benzoic acid put in the solution. I know I could easily find the moles using the mass and the formula weight but the question asks me to use the molality and mass in kg of the lauric acid to calculate the moles of benzoic acid solute.

[enahs]

So you are saying there is 1 gram of benzoic acid in how much lauric acid?

Type the question out word for word, starting over.

[chay722]

There is 1 g benzoic acid put in 8 g lauric acid. The calculated molality of the solution is .06 mol/kg. Use the molality and the mass in kg of lauric acid solvent to calculate the moles of benzoic acid solute.

[enahs]

1g/122g/mol/0.008kg = 1 Molal


Are you sure you originally calculated the molality correctly?