ƒ¢Tf = Kf * m
Where ƒ¢Tf = Depression in the freezing point
m = molality
Kf = molal depression constant
For water it is 1.86 0C/ m
ƒ¢Tf = Tpure - Tsolution
= 0 0C - ( --23.30C)
m = no.of moles / mass of solvent
= mass in g of solute / 62.068 g /mol * 9.6 Kg ( since 1L = 0.96 Kg )
23.30C = 1.86 0C/ m * mass in g of solute / 62.068 g /mol * 9.6 Kg
ˆ mass in g of solute = 23.30C * 62.068 g /mol * 9.6 Kg / 1.86 0C/ m
.........................i have a lab on monday , i am getting ... 7464.17g.
SO the number is suppose to be that high, i guess i will have to check my significant digits and just go over it once more to see what i get. Thanks
ya i guess so ..can u do me favour plz ... can u plz explain ur answer in detail i didn't either get it .. coz i copy paste from some other website.. plz can u explain me how to do this question .. sine i need to learn , in order to be able to rite my quiz lol ..ok thnks really appreciate thnks
bye..hope u ll reply asap thnks in advacne
Hey im not really sure, but i had posted it up before and someone helped me here, but i will try my best:
The formula you'll need to use for this problem is:
Freezing Point = Kf * m
Start by calculating the Freezing Point:
=-23.0 degrees C
Next, use the formula to calculate the target colligitive molality:
= -23=1.86 * Cm
= 12.37= Cm
Finally, we determine the number of grams needed to produce a solution with a Cm of 12.37
Note that colligitive molality = i x molality
In C2H602, i = 1.
Since we really need to solve for molality not colligitive molality, we divide that Cm through by 1 to simplify our calculation.
It remains the same because in this case the The i for ethylene glycol is 1, but if it was some other number you would have the divide by that number:
Hmm If you want to know how to find out i its on page 556 in the textbook:
So now we use the eqaution
molality= moles of solute (KNO3)/ kg of solvent (water)
We know that we have 10.0L of water and that water has 1g/mL (by its density) so , the kg would be the same as the litres:
12.37m= moles C2H6O2/ 10.0kg water
123.7= moles of C2H6O2
NOw i just take the moles and times by molar mass to get the grams of C2H6O2 that are being used
I think both of you have made arithmetic/copying errors. First, I have molar mass ethylene glycol as 62.068. The problem says -23 degrees and not 23.3. And I didn't round the molality; I just left the number in my calculator.
delta T = i*Kf*m
23 = 1*1.86*m
Solve for m.
Now you can solve for this is several steps BUT I've worked with this stuff long enough that I have put all of it together. So, in detail,
molality = mols/kg solvent
You know molality and kg solvent (by the way that will be 10 kg and not 9.6 kg), solve for mols. Then we know mols = g/molar mass. You know molar mass and mols, solve for grams. BUT you can put all of together to make a neat formula that is
kg solvent x molality x molar mass = grams.
10 x (23/1.86) x 62.068 = grams
I get 7,675 grams. I expect you should put that in scientific notation because 23 have only two significant figures in the 23 or 3 in the 1.86. I don't know how many zeros you omitted in the problem (23 or 23.00 and ensity 1 or 1.00 etc). You work the number of s.f. out depending upon your actual problem numbers.
sorry i made an error in the question it is 23.3 , but thank you so much i will work on the question
holy, we CHMA11 kids are all over the internet. trying to find how to work out the pre-labs and mastering chemistry homework.
omg yeah, its the only source of help at utsc -_-
anyone remember any questions for the lab quiz?
oh my God ...how did you guys do at the mid-term.. lol i beat the class average by 15% which is great, I guess
WHATS 3HO= 4HO
the write way to write it in scientific notation is 778*10^1, because there is three s.f. in 23.3 and as well in 1.86.