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### Topic: Chemistry Help  (Read 11474 times)

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#### charles741

• Guest ##### Chemistry Help
« on: February 06, 2005, 11:54:53 AM »
I spent the last week trying to figure these questions out, but I just can't seem to get the right answer. I was hoping someone can help me out. Here's the questions:

1)Calculate the osmotic pressure of this solution: 69.7 g of a solute with a molecular mass of 2790 g are dissolved in enough water to make 1.00 dm3 of solution at 20 degrees C. What is the osmotic pressure of the solution in kPa?

2)Calculate the osmotic pressure of this solution: 18.6 g of a solute with a molecular mass of 8940 g are dissolved in enough water to make 1.00 dm3 of solution at 25 degrees C. What is the osmotic pressure of the solution in atmospheres?

3)Calculate the molecular mass of the nonionic solutes: 64.3 g of solute in 390 g of water raises the boiling point to 100.680 degrees C.

4)Calculate the molecular mass of the nonionic solutes: 8.02 g of solute in 861 g of water lower the freezing point to -0.430 degrees C.

5)Compute the freezing point of this solution: 25.5 g C7H11NO7S (4-nitro-2-toluenesulfoinoic acid dihydrate) in 100 g H2O. (nonionizing solute)

6)Find the vapor pressure of a water solution in which the molefraction of HgCl2 (a nonvolatile solute) is 0.163 at 25 degrees C. What is the vapor pressure in kPa?

Here's what I got for the answers but I'm told they're all wrong:
1)0.1 kPa
2)0.05 atm
3)4834.6 g
4)34.9 g
5)-1.9
6)0.63 kPa

Thanks for all the help.
« Last Edit: February 06, 2005, 06:11:27 PM by charles741 »

#### savoy7

• Guest ##### Re:Chemistry Help
« Reply #1 on: February 06, 2005, 02:04:30 PM »
Yep

You want molar mass of solute - that means you need g/mole

I will also tell you that water's boiling pt / freezing pt constants are:  0.52 C/molal
and 1.86 C/molal

Why did they have to tell you that it was a nonionic solution?  This is very important.  I'll cover just the basics here:  in order for water to boil it's molecules have to have enough energy to break it's surface tension and become a gas.  Now if you throw some solute in there, they (solute particles) get in the way of the water and it requires more energy to get the water to boil.  Thus the boiling pt increases.  In order for water to freeze the molecules have to get closer together (I don't want to get into the structure for ice & the whole density thing), but let's just say that the water has to get closer together.  If there is solute there, they get in the way.  So more energy has to be lost, before the water will freeze.  Thus, the freezing pt of the solution is lowered.

Both Ideas deal with the number of particles of solute in the solution.  You have to know if the solute will break apart or stay together.  Ionic substances will disassociate and form more particles than a molecule (covalently bonded & won't disassociate).

With this in mind you need to know if the solute is ionic or molecular, how to determine molality and freezing/boiling pt depression/elevation.

show some work on these questions and post again, I'm sure someone will help.
here's a website that might:  http://dbhs.wvusd.k12.ca.us/webdocs/Solutions/BP-Elev-and-FP-Lower.html

#### charles741

• Guest ##### Re:Chemistry Help
« Reply #2 on: February 06, 2005, 06:09:16 PM »
Here's the work and answers that I got from the questions, if it's possible can you tell me what I did wrong, here's the work and answers:

1)69.7g * (1/2790) = 0.02498 M

R=0.0821 L*atm/mol*K

T=20 C + 273.15=293 K

TT= MRT = (0.02498 mol/L) * (0.0821 L*atm/mol*K) * 293 = 0.6 atm

0.6 atm * (0.1333 kPa/1 atm) = 0.1 kPa

2)18.6g * (1/8940) = 0.0021 M

T = 25 C + 273.15=298 K

TT= MRT = (0.0021 mol/L) * (0.0821 L*atm/mol*K) * 298 = 0.05 atm

3)0.680 = 0.512/molal * M

0.680/0.512 = 1.33 M

(64.3/X) * (1.33/100)

X=4834.6

4)0.430 = 1.86/molal * M

0.430/1.86 = 0.231 M

(2.02/X) * (0.231/100)

X=34.9

5)25.5 g * (1/253.231) = 0.1007

0.1007/0.1 = 1.007

1.86 * 1.007 = -1.9

6)0.163 * 28.8 = 4.7 torr

4.7 torr * (0.1333 kPa/1 torr) = 0.63 kPa

#### savoy7

• Guest ##### Re:Chemistry Help
« Reply #3 on: February 06, 2005, 08:06:04 PM »
I got a little time - watching the game.

I'll help you with 3 - (4&5) are related

you want the molar mass of the solute = g/moles

you already know the mass - 64.3 grams so you need to find the moles of solute

Since this is nonionic you don't have to worry about a disassoication factor.
You calculated the molality = 1.31 m using bp elevation -
that's right
molality = moles of solute/kg of solvent
you are given the mass of the water - being 390 g  - convert to kg
You should be able to use that info to determine the moles of solute

then just 64.3 g/moles and that's it

this was your error (64.3/X) * (1.33/100)

the 1.33/100  - I don't know what you were doing here?

for an answer - I got ~ 126 g/mole without a calculator

check it out-never trust someone else's work - this should help with 4&5
gotta go NE is driving
Savoy out

#### savoy7

• Guest ##### Re:Chemistry Help
« Reply #4 on: February 06, 2005, 08:31:17 PM »
Back - halftime

Here's your problem 1 -
1)69.7g * (1/2790) = 0.02498 M

R=0.0821 L*atm/mol*K

T=20 C + 273.15=293 K

TT= MRT = (0.02498 mol/L) * (0.0821 L*atm/mol*K) * 293 = 0.6 atm

0.6 atm * (0.1333 kPa/1 atm) = 0.1 kPa

Looks good except the last step - 101.325 kPa/1atm
should get around 60.9 kPa

#### charles741

• Guest ##### Re:Chemistry Help
« Reply #5 on: February 07, 2005, 07:00:20 PM »
Thank you, Savoy7.  With the help you gave me I finally got the questions right.