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### Topic: Osmotic pressure and Freezing point help please!  (Read 2282 times)

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#### xstellar1x

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##### Osmotic pressure and Freezing point help please!
« on: May 11, 2010, 02:09:15 PM »
I am taking a chemistry course online at 40, so I can move on to Anatomy for a nursing degree...ugh...never had it in high school so I am struggling...I need help with this question...

18.6 grams of a solute with molecular mass of 8940g are dissolved in enough water to make 1.00 dm3 of solution at 25c. What is the osmotic pressure...to the nearest tenth.

I get .05 atm...but this is not the answer....

18.6 / 8940 = 0.00208 mole solute

Molarity = 0.00208 / 1 L = 0.00208

T = 2 + 273 = 275 K

osmotic pressure = 0.00208 x 0.0821 x 275 = 0.0470 atm

2)  25.5 g C7H11NO7S 4-2toluenesulfonoic acid dihydrate in 1.00 x 10^2 g H2O (non ionising solute) What is the freezing point to the nearest tenth...

I get -1.87 which I round off to -1.9 but this is not right either....

I would appreciate any help

#### opti384

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##### Re: Osmotic pressure and Freezing point help please!
« Reply #1 on: May 14, 2010, 09:42:43 AM »

Well for the first question, I think there is a problem with the temperature. The temperature should be in Kalvin, therefore it should be 298 (273+25)K not 275K. Check the answer after you plug in the right temperature. If this doesn't do the trick, you should check the volume of the solution. It says 1.00 dm^3 which is 100L.

For the second question, you should consider the number of particles existing in the solution. For example, when 1m of NaCl is dissovled in water, the decrease of freezing point is twice of 1m of molecules that do not ionize such as sugar. In this case, since the compound is an acid you should google it and find how many particles will exist due to the ionization.