Molality/Molar Question.....HELP!

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Matty44

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A solution of vinegar is 0.763M acetic acid (HC2H3O2). The density of the vinegar(the solution) is 1.004 g/mL. What is the molal concentration of acetic acid??? I need help! got a test tomorrow and this practice problem is buggin me. Thanks in advance.

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is the answer 7.6 x10^-4?

I just did .763 moles/L x 1L/1004g = .0007599 mol/kg
 
I don't know the answer. Its just a question on a review sheet my teacher gave us. Thanks for the response, but is there anyone else who KNOWS FOR SURE how to do this?
 
A 0.763 M HAc= 0.763 moles/liter.
Since 1 liter will weight 1.004Kg then

0.763/1.004Kg= xMolal/1.0Kg
x=0.7599 Molal
 
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is the answer 7.6 x10^-4?

I just did .763 moles/L x 1L/1004g = .0007599 mol/kg


It should be .763 mol/L * 1L/1.004kg = .763 mol/kg

*Please note that 1.004 g/mL = 1.004 kg/L*
Also, molality is moles/kg not moles/g

Edit: Oops, the doc beat me to the punch.
 
Thanks for all the responses guys. I get the math portion, but i'm still having a bit of trouble conceptualizing how the numbers fit in to the real world. Also, with sig figs, shouldn't it be .760mol/kg?
 
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