What mass of ethylene glycol C2H6O2, the main component of antifreeze, must be added to 10.0 L water to produce a solution for use in a car's radiator that freezes at -23C? Assume density for water is exactly 1g/mL.
First we find out the freezing point:
=(-23.0)-(0.0_
=-23.0 degrees C
Calculate target Colligative Molarity:
= -23=1.86 * Cm
= 12.37= Cm.
i* molality
i= 1
12.37= 1*m
12.37= m
molality= moles of solute(C2H6O2)/kg of solvent water
12.37m= moles C2H6O2/ 10.0kg water
123.7= moles of C2H6O2
123.7moles C2H6O2 * 62.26g C2H6O2
= 7701.56 g of C2H6O2
I think this number is quite high can someone please double check to see if i got the answer right.
Thank YOu
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
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.
Formula :
ĢTf = Kf * m
Where ĢTf = Depression in the freezing point
m = molality
Kf = molal depression constant
For water it is 1.86 0C/ m
ĢTf = Tpure - Tsolution
= 0 0C - ( --23.30C)
= 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
= 7464.17g
.........................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
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:
Freezing Point
(-23.0)-(0.0)
=-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.
12.37= 1*m
12.37= m
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
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.
Bot, Please answer this question
Find the mean absolute deviation (MAD) of the data
98 132 295 68 176 89 287 198 94 68 240 94 369 94 155