Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: student8607 on December 01, 2008, 07:27:07 AM
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I'm a little confused at how I am supposed to set this up:
The density of a 0.258m solution of glucose in water is 1.073g/mL. What is molarity?
molarity = mol solute / L solution
molality = mol solute / kg solvent
sucrose's molar mass is 180.2g
density = mass/volume
so would the molarity have a volume of 0.001L?
then just convert grams to moles? 0.00564mol
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Dear student8607;
Are you sure, that the question asks for molaRity and not molaLity?
(I think for molaLity.)
Good Luck!
ARGOS++
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yes
then there is a similar question below that asks for molality
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Dear student8607;
In this case for molaRity: What is 0.258m solution meaning?
Good Luck!
ARGOS++
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perhaps that there are 0.258moles of solute for every 1 kg of solvent?
so now we have moles for molarity :D
just need volume...can we use the 0.01L?
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Dear student8607;
That’s not correct! (http://not correct!)
0.258m solution means contrary, 0.258 molaR (= moles / Liter) !!
Do you see now, that the question for results in molaRity makes no sense!
(But for molaLity!)
Good Luck!
ARGOS++
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but i thought little m was for molality (moles/kg)
big m (M) is for the molarity?
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Dear ;
Sorry!, - so it’s my mistake for that!
But then it’s similar to your other question! Solve it first.
Sorry again!
Good Luck!
ARGOS++
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hmm
D = m/v
0.258m = 0.258mol solute per 1 kg of solvent
molar mass for glucose is 180.2g/mol
0.258mol = 46.5g
D = m/v
1.073g/mL = (46.5g) / (?L)
43L? ???
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Dear student8607;
The first part is ok.
But to turn to the opposite you have to ask in how much masstotal are the 0.258 moles contained, and second: what is the volume of the masstotal?
Then you should be able to answer the question for molaRity.
Good Luck!
ARGOS++
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Can we just assume 100mole or something?
Sorry, I'm not seeing the connection yet.
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Dear student8607;
Do you remember?:
More important for you is: masstotal = mass Solute + mass Solvent!
(I think you know how to use it in mass%.)
Good Luck!
ARGOS++
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ok so if i assume 0.258molality means 0.258moles (per every 1 kg of solvent)
then i can convert that to grams (mass) = 46.5g of SOLUTE
but how am i to get mass of water for mass %
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Dear student8607;
I told you already that mass of Solute (46.5g = 0.258 moles) is correct.
But what is the mass of the Solvent? – You wrote it in the parentheses from the definition.
(You don't need any %!)
Good Luck!
ARGOS++
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46.5g - solute
1000g(aka 1kg) - solvent
1046.5g - total for solution
M = mol solute / L of solution
M = 0.258mol / ?L
D=m/v
1.0173g/mL = (1046.5g) (?mL)
0.000972mL???
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Dear student8607;
Please control your last calculation; you did a BIG comma mistake!
Use the Units for control!
And if you have corrected it, then calculate the final molaRity from it.
Good Luck!
ARGOS++
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yeah i see that my calculations has way too many zeroes
but i doubled checked
density was given in g/mL
and we said 1kg for solvent. so that's 1000g + the 46.5g of solute from the molality
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Dear student8607;
Please control again! (masstotal is ok!)
But I got: 1046.5g / 1.0173 g/ml = 1028.7 ml (= voltotal)
(If density is finally correct.)
Good Luck!
ARGOS++
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oh you are right
it is mass over density not the other way around
so then 0.2588mol/1.028L = 0.251M ;)
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Dear student8607;
You did it! - Hard but ultimately!
Repeat the small conversion once again, to “get” really behind it!
Good Luck!
ARGOS++
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YAY - thanks very much
so i got bold and tried another problem (molarity to molality)
and i failed >:(
The density of a 0.500M solution of acetic acid in water is 1.0042g/mL. What is the molality?
Molar mass is 60.05g
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0.5mol acetic acid = 30.025g of acetic acid
D times V = total mass
(1.0042g/mL)(1000mL from the 0.500M) = 1004g total
so i figured 1004-30.025 would be grams of solvent (0.974kg)
0.5mol / 0.974kg = 0.513?
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oh wait, my fault, that WAS correct
0.513 is right
yay :o
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Dear student8607;
Correct!
Good Luck!
ARGOS++
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Examples of general approach:
http://www.chembuddy.com/?left=concentration&right=percentage-to-molarity
http://www.chembuddy.com/?left=concentration&right=percentage-to-molality
And ultimate approach:
http://www.chembuddy.com/?left=CASC&right=concentration-cheat-sheet