You have a container of 17.42 M conc. acetic acid (CH3COOH(aq)) with density 1.05 g/mL, find the molality of a 150ml sample of the solution. molality = moles solute/Kg solvent moles solute = (0.150 L)(17.42 mol/L) = 2.613 mol CH3COOH molar mass of CH3COOH = 60.06 g/mol grams of solution = (150mL)(1.05 g/mL) = 157.5 g solution grams of solute = (2.613 mol)(60.06 g/mol) = 156.9 g CH3COOH so the total grams of water in this solution would be: g H2O = g solution - g CH3COOH = 157.5g - 156.9 g = 0.6g H2O 0.6g H2O = 0.0006 Kg H2O so molality = (2.613 mol CH3COOH) /(0.0006 kg H2O) = 4355 molal this seems rather high to me but if its 'concentrated' acetic acid, I know that you can get it to about 98% as glacial acetic acid so perhaps there is 4355 moles of acetic acid per kg of water. Did I do this correctly? I know that typically the solvent is the larger component of a solution but the fact that it gives me CH3COOH(aq) would imply that H2O is the solvent still in this situation, I have notes from class where we did the same thing with 12M HCl and took a 750ml sample that turned out to be 16.2 molal but the sample was much larger and the concentration was smaller so I don't know if this is way too big of a number or not. Can anyone verify that I have solved this correctly? The main problem I am having is that if you take a full liter sample you get about half the molality that you do for the 150mL sample, is this normal? by this I mean I did: (1L)(17.42M) = 17.42 mol acetic acid molar mass = 60.06 g/mol mass acetic acid = (17.42 mol)(60.06 g/mol) = 1046 g acetic acid mass of solution = (1000 mL)(1.05 g/mL) = 1050 g solution mass of solvent = mass solution - mass acetic acid which is: 1050g - 1046g = 6g solvent (H2O) = 0.006 Kg solvent so molality = 17.42 mol acetic acid/ 0.006 Kg H2O = 2903 molal I thought that molality was not supposed to change? Or am I just looking at this problem completely wrong?
You have a container of 17.42 M conc. acetic acid (CH3COOH(aq)) with density 1.05 g/mL, find the molality of a 150ml sample of the solution.
molality = moles solute/Kg solvent
moles solute = (0.150 L)(17.42 mol/L) = 2.613 mol CH3COOH
molar mass of CH3COOH = 60.06 g/mol
grams of solution = (150mL)(1.05 g/mL) = 157.5 g solution
grams of solute = (2.613 mol)(60.06 g/mol) = 156.9 g CH3COOH
so the total grams of water in this solution would be:
g H2O = g solution - g CH3COOH = 157.5g - 156.9 g = 0.6g H2O
0.6g H2O = 0.0006 Kg H2O
so molality = (2.613 mol CH3COOH) /(0.0006 kg H2O) = 4355 molal
this seems rather high to me but if its 'concentrated' acetic acid, I know that you can get it to about 98% as glacial acetic acid so perhaps there is 4355 moles of acetic acid per kg of water.
Did I do this correctly? I know that typically the solvent is the larger component of a solution but the fact that it gives me CH3COOH(aq) would imply that H2O is the solvent still in this situation, I have notes from class where we did the same thing with 12M HCl and took a 750ml sample that turned out to be 16.2 molal but the sample was much larger and the concentration was smaller so I don't know if this is way too big of a number or not. Can anyone verify that I have solved this correctly?
The main problem I am having is that if you take a full liter sample you get about half the molality that you do for the 150mL sample, is this normal?
by this I mean I did:
(1L)(17.42M) = 17.42 mol acetic acid
molar mass = 60.06 g/mol
mass acetic acid = (17.42 mol)(60.06 g/mol) = 1046 g acetic acid
mass of solution = (1000 mL)(1.05 g/mL) = 1050 g solution
mass of solvent = mass solution - mass acetic acid which is:
1050g - 1046g = 6g solvent (H2O) = 0.006 Kg solvent
so molality = 17.42 mol acetic acid/ 0.006 Kg H2O = 2903 molal
I thought that molality was not supposed to change? Or am I just looking at this problem completely wrong?
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