[mork]

Here are the questions I just cannot figure out..

#1. A bottle of rum contains 50% ethanol (c2h5oh), by volume.  The density of ethanol is 0.789 g/mL and that of water is 1.00 g/mL.  What is the freezing point of this rum?

Waters Kf = 1.86 degrees C/m (molality) 
ethanol's molecular weight is 46g
kf = freezing point constant



#2.  Estimate the Osmotic Pressure associated with 32.5g of an Enzyme of molecular weight 4.21 X 10^6 dissolved in 2000mL of   water at 35 degrees Celsius.

Osmotic pressure is -  pi=(n/V)RT   R= 0.0821 L x ATM / mol X Kelvin 
also you convert the 35 degrees celsius to kelvin 35+273 = 308 degrees Kelvin
Turn the 2000 mL into Liters= 2L
So V=2L  T= 308 K 
I did some weird stuff to figure this out and I think it is wrong the answer i got is 9.76 X 10^-5 ATM I did this by:

32.5g/4.21 X 10^6 g/mol = 7.72 X 10^-6
now n= 7.72 X 10^-6

pi= (7.72 X 10^-6 mol / 2L)(0.0821 L X ATM/ mol X Kelvin)(308 K) = 9.76 X 10^-5 ATM

[Albert]

Question number 2 is correct.

For what concerns #1, you might find this quick formula very useful:

molality = mol/kg = (0.789g/mL x 50mL/100 mL x 10⁺³mL/kg)/ 46g/mol

[mork]

okay, well i have written that down but im having a problem because all of the examples i have actually HAVE temperatures given.
im trying to see what i can do with the molality of ethanol now gimme a few mins to see if i can get something

[Albert]

What is the freezing point of pure water? 

[mork]

Okay heres what I got, I think im close but im off track somewhere..

molality of ethanol = (0.789 g/mL)(0.7x10^3)= 0.552x10^3

Change in Tf= (1.86 degrees C/molality)(0.552 x 10^3 molality of ehtanol) = 1.03 x 10^3 degrees C

then i think it freezes at a temperature BELOW the freezing point of pure water so
0.00degrees C - 1.03 x 10^3 degrees C   =   -1.03 x 10^3 degrees C is the freezing point..

I think that is wrong though because i doubt that it freezes at -1030 degrees celsius lol thats REALLY cold..

i think my idea is right but 0.552 x 10^3 is the molality of ethanol and not the solution and i think i need to somehow figure out the molality of the solution THEN put that where i have the molality of ethanol in the below equation

Change in Tf= (1.86 degrees C/molality)(0.552 x 10^3 molality of ehtanol) = 1.03 x 10^3 degrees C

[Albert]

The concentration you got from the equation I showed you is in g/L: you must then convert it into moles.

I've now edited the equation so that it can be used to calculate the molality of ethanol directly.

[mork]

I figured it out w00t w00t!!!!

okay here is how i got the molality (m) of ethanol:

moles ethanol/1000g water = 1 mole ethanol/46g ethanol (molecular weight) X 0.789g ethanol/1mL X 500mL ethanol/500mL H2O X 1000mL H20/1000g H20 = 17.5m so that is the molality of ethanol

now I used the Kf of water and the m of ethanol to get the answer here:

The change in the freezing point temperature (Tf)= (the Kf of water) (m of ethanol)= answer

Tf= (1.86 degrees C/m)(17.15m)= 31.9 degrees celsius

[mork]

I got a couple other actual easy questions here that im pretty sure i got right just kinna got a small doubt but here goes...

#3. An ideal gas has a volume of 10.0L and a pressure of 200 torr. what is the pressure if the volume is expanded to 25.0L? assume constant n & t

Boyles Law: PV=PV
convert Torr to ATM: 1atm/760 torr X 200 torr = .263 ATM

(.263 ATM)(10L)= (P)(25L)= 0.105 ATM

my question about the above problem is if I associate the right pressure with the right volume.. like should it be (.263 ATM)(25L)=(P)(10L)= I ask this because in one of my example problems that the teacher worked on the board a while back had the pressure and volume mixed up (at least it seemed to be to me) and I am hoping that I have it right.. to me it seems common sense that .263 ATM would go with the 10L but thats now how one of my examples are, but maybe the teacher screwed up in that example.. think its okay?



#4. How many molecules of He are present in a container that has a volume fo 10.0L at a pressure of 76.0 torr and a temperature of 15 degrees C? assume ideality.

IDEAL GAS LAW: PV = nRT R= 0.08206 L X ATM / mol X degrees K
converted 76 torr to .1ATM

PV=nRT
(0.1 ATM)(10 L)= (n)(0.08206 L x ATM / mole X Kelvin)(288 degrees Kelvin)
n = 0.0423 moles HE

now I have the moles of HE but I need molecules of He heres where I actually may have messed up. in order to convert to molecules of He here is what i did

I used avogadros number 6.022 x 10^23 molecules

molecules He= 6.022 X 10^23 molecules He/1 mole He X 0.0423 moles of He = 2.547 X 10^22 molecules of He

my doubt is that maybe i shouldnt have used avogadros number but i dont no another way to get the molecules, i think its right though hehe










also on the second question i have the osmotic pressure one, are you sure that ones right, i still have my doubts about that one hehe