Thursday

January 19, 2017
Posted by **Saira** on Saturday, January 24, 2009 at 1:06pm.

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

- CHemistry.... -
**Casandara**, Saturday, January 24, 2009 at 1:11pmFormula :

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)

= 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. - CHemistry....Please double check DrBobb -
**Saira**, Saturday, January 24, 2009 at 1:19pmSO 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

- CHemistry....Please double check DrBobb -
**Casandara**, Saturday, January 24, 2009 at 1:28pmya 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 - CHemistry....Please double check DrBobb -
**Saira**, Saturday, January 24, 2009 at 1:51pmHey 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 - CHemistry....Please double check DrBobb -
**DrBob222**, Saturday, January 24, 2009 at 2:20pmI 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. - CHemistry....Please double check DrBobb -
**Saira**, Saturday, January 24, 2009 at 2:28pmsorry i made an error in the question it is 23.3 , but thank you so much i will work on the question

- CHemistry....Please double check DrBobb -
**ur class fella**, Monday, February 2, 2009 at 7:21pmholy, we CHMA11 kids are all over the internet. trying to find how to work out the pre-labs and mastering chemistry homework.

- chem -
**haha**, Tuesday, February 3, 2009 at 7:44pmomg yeah, its the only source of help at utsc -_-

- CHemistry....Please double check DrBobb -
**eric**, Tuesday, February 3, 2009 at 8:11pmanyone remember any questions for the lab quiz?

- CHemistry....Please double check DrBobb -
**ur class fella**, Tuesday, February 17, 2009 at 12:53amoh my God ...how did you guys do at the mid-term.. lol i beat the class average by 15% which is great, I guess

- CHemistry....Please double check DrBobb -
**LANE**, Monday, March 16, 2009 at 10:29pmWHATS 3HO= 4HO

- CHemistry....Please double check DrBobb -
**ameer**, Monday, January 25, 2010 at 10:15pmthe 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.