[xstellar1x]

I am taking a chemistry course 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....

the work:
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

[Grundalizer]

Well I see in number 1 that you said the temp needed to be 25C (room temp) but in your calculation you only added 2 to 273K to get 275K when it should in fact be 298K

[MrTeo]

1) Why do you use 275 K (273+2)? From the data you've provided looks like the temperature is 298 K (273+25)...

2) $$ \Delta T=K_{cr} \cdot m /$$, try to revise your calculations or post your attempts  

[xstellar1x]

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....

the work:
18.6 / 8940 = 0.00208 mole solute

Molarity = 0.00208 / 1 L = 0.00208

T = 25 + 273 = 298 K

osmotic pressure = 0.00208 x 0.0821 x 298 = 0.0508 atm isn't that still .05 the nearest tenth... ?? I am lost...sorry;)

??

[sjb]

What units do your equation need to be in?

How many significant figures?

[Grundalizer]

I'm going to give you a tip I just learned last year (and I'm going to be a senior in college this coming fall!) it's called dimensional analysis, as sjb was hinting to.  Now I know you know the answer and formula in this case, but if you are ever lost, and you KNOW what the units of your answer are, you can arrange the units you DO have in such a way as they cancel out to give you the units you need.

So even though it's a bit of a pain to write out units over the internet, and on paper, DO it.  I used to never write units out, and just work the numbers in my calculator or head, but as the problems get more complex, you WILL lose track and have to keep starting from the beginning.

As for Number 1, lets rework it.

you have 18.6g of a solute with molecular mass of 8940g, dissolved in 1L at 25C.

18.6g / 8940g = 2.08 x 10^-3 mol/L

Now,
 Where;
M is the molar concentration of dissolved species (units of mol/L).
R is the ideal gas constant (0.08206 L atm mol-1 K-1, or other values depending on the pressure units).
T is the temperature on the Kelvin scale.

We know the temperature is 298K (273K+25K)
We know the ideal gas constant given above.
We know the molar concentration of the solute.

So Osmotic pressure=(2.08 x 10^-3 mol/L)(0.08206 L atm mol-1 K-1)(298K) = 0.0509 atm (3 sig figs)= 0.051 atm = 0.1 atm (to the nearest tenth)

0.05 atm is to the hundreth, not tenth

[Grundalizer]

As for #2.

ΔT_f = K_f · m_B will be used later for freezing point depression.

Where  :delta: T is the change in temperature, K is a constant based on the solvent, and m the MOLALITY (not to be confused with molarity) of the solution.  Molality is moles of solute/Kilogram of solvent. 

So, first lets figure out how many kg of solvent we have.

We are given  1.00 x 10^2 g H2O (non ionising solute) which = 2g of water which = 0.002kg water. 1L=1kg for water. That acid does ionize in water...but I guess in this case they want us to NOT ionize it because they list (non ionising solute).  Just for the record, the more ions the solute turns into, the greater the freezing point depression, which is why Calcium Chloride (CaCl₂ is used on highways to de-ice instead of regular NaCl (one turns into 3 ions, the other 2)).  But anyway.

So we now know that 2g of water = 0.002 kg of water, which in this case is our kg of solvent.

The acid has a molecular mass of 172.20 g/mol and we have 25.5 g.

Recalculate the molar mass and tell us what you get.

Also, the K_f value for water is 1.86 C/m  (Degrees C per molal)

Then plug the molality and the Kf into the equation I linked above and tell us what you get and if it's correct.

[sjb]

I'm going to give you a tip I just learned last year (and I'm going to be a senior in college this coming fall!) it's called dimensional analysis, as sjb was hinting to.  Now I know you know the answer and formula in this case, but if you are ever lost, and you KNOW what the units of your answer are, you can arrange the units you DO have in such a way as they cancel out to give you the units you need.

Well, that wasn't quite what I was trying to get at. You're not wrong, though.

I don't know whether the answer should be in pascals, atm, mmHg, ...? I think the answer of 0.0508 atm is right as far as it goes, but I suppose the amount with the most error is the volume, so 0.05 atm is probably the answer iff it's required in atm.

[Grundalizer]

True, we never found out what units pressure had to be in...as there are many different values of R.

[xstellar1x]

Ok..SJB gave me an ah ha moment so the answer to #1 is .1atm

working on #2

[Grundalizer]

Sure, give him all the credit

[xstellar1x]

Grundalizer~You were extremely helpful too!! I just read his post first....Still can't get problem 2, I must be totally dumb...

[Grundalizer]

25.5g of the acid with a Molar mass of 172.2 g/mol.

That means we have 0.148 mols of the acid.

ooo and just noticed its 100g of water, not 2.... I don't know what I was thinking, just missread the exponent.
 So now we have 0.100 kg of solvent

0.148 mol acid / 0.100 kg water = 1.48 m (molal)

(1.48 molal) * (1.86 C/molal) = 2.75 C

So the freezing point will drop by 2.75 degrees Celcius