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Feb 11 2014 12:51am
i just did like 30 of these last week and havent slept its late im brain dead

At 300 K a 500.0 mL solution containing 0.4314 g of dextrose has an osmotic pressure of 89.6 mm Hg. What is the molar mass of dextrose?

osmotic pressure = 89.6 mmHg
Molarity= x
gas constant .08206
temp kelvin 300

osmotic pressure = MRT

89.6= M (.08206) (300)

M= 3.64

I am totally forgetting how to find the molecular weight from the molarity.
Something seems off.
3.64 mole dextrose per 1 Liter solution. Since we only have .5 L, do I divide the moles as well? So, 1.82 mole dextrose = .4314 g dextrose. No way thats right..

i copied it exactly but im thinking something may be off in the problem?

This post was edited by fingerling on Feb 11 2014 01:13am
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Feb 11 2014 07:53am
Quote (fingerling @ Feb 11 2014 12:51am)
i just did like 30 of these last week and havent slept its late im brain dead

At 300 K a 500.0 mL solution containing 0.4314 g of dextrose has an osmotic pressure of 89.6 mm Hg. What is the molar mass of dextrose?

osmotic pressure = 89.6 mmHg
Molarity= x
gas constant .08206
temp kelvin 300

osmotic pressure = MRT

89.6= M (.08206) (300)

M= 3.64

I am totally forgetting how to find the molecular weight from the molarity.
Something seems off.
3.64 mole dextrose per 1 Liter solution. Since we only have .5 L, do I divide the moles as well? So, 1.82 mole dextrose = .4314 g dextrose. No way thats right..

i copied it exactly but im thinking something may be off in the problem?


e/ sorry read wrong early morning -.-

your error is using the pressure in mmHg while the gas constant R that you are using is in atm. convert mm Hg to atm and you should be fine and get an answer around 180.16 :)

This post was edited by cialda on Feb 11 2014 07:57am
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Feb 11 2014 11:53am
Quote (cialda @ Feb 11 2014 08:53am)
e/ sorry read wrong early morning -.-

your error is using the pressure in mmHg while the gas constant R that you are using is in atm. convert mm Hg to atm and you should be fine and get an answer around 180.16 :)


Ah dang you're right

Thanks
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Feb 11 2014 02:13pm
What is the value of the van't Hoff factor for KCl if a 1.00m aqueous solution shows a vapor pressure depression of 0.734 mmHg at 298 ∘C? (The vapor pressure of water at 298 K is 23.76 mmHg.)


could someone help with this one as well? thanks
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Feb 11 2014 07:56pm
Quote (fingerling @ Feb 11 2014 03:13pm)
What is the value of the van't Hoff factor for KCl if a 1.00m aqueous solution shows a vapor pressure depression of 0.734 mmHg at 298 ∘C? (The vapor pressure of water at 298 K is 23.76 mmHg.)


could someone help with this one as well? thanks




anyone
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Feb 11 2014 09:09pm
Quote (fingerling @ Feb 11 2014 07:56pm)
anyone


is the "m" molality or molarity?

also what level class is this? in an ideal solution, the van't hoff factor should just be 2 iirc.

for non ideal, i believe it should be:

(Pnot - P)/Pnot = i*mole fraction
Pnot = vapor pressure pure solvent
P = vapor pressure of the resulting solution
i = van't hoff factor
mole fraction = mole solute/total mole

This post was edited by cialda on Feb 11 2014 09:17pm
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Feb 11 2014 10:01pm
Quote (cialda @ Feb 11 2014 09:09pm)
is the "m" molality or molarity?

also what level class is this? in an ideal solution, the van't hoff factor should just be 2 iirc.

for non ideal, i believe it should be:

(Pnot - P)/Pnot = i*mole fraction
Pnot = vapor pressure pure solvent
P = vapor pressure of the resulting solution
i = van't hoff factor
mole fraction = mole solute/total mole



(Pnot - P)/Pnot = i*mole fraction
Pnot = 23.76
P = 23.76 - 0.734
i = unknown
mole fraction solute = mole solute/total mole

1 molality = 1 mole KCl / 1000 g H2O
1000 g H2O = 55.56 mole h2o
mole fraction = 1/(55.56+1) ~= 0.0177

((Pnot - P)/Pnot)/mole fraction = 1.747


another way to do it:

Mole fraction solute calculated (theoretical) = 1/56.56 = 0.0177

P = Pnot *mole fraction solvent
P/Pnot = 0.969 = mole fraction solvent
1 = mole fraction solvent + mole fraction solute
mole fraction solute = 1 - mole fraction solvent = 0.0309

i (vant hoff factor) = observed colligative value / calculated value with no association/dissociation
i = 0.0309 / 0.0177 = 1.747

hopefully this helps. not too familiar with these formulas myself.

This post was edited by cialda on Feb 11 2014 10:03pm
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