Don't know why, but I keep getting stuck on what WA says is the wrong answer for these two problems. anyone care to assist on 2 fairly easy problems?


How many grams of K3PO4 (Mm=212 g/mol) must be added to 655 g of water to make a solution with a colligative molality of 0.165 m?
? g K3PO4

A solution is found to be 0.1161% by weight CoF3. What is the concentration of the F1- ion expressed in ppm?
? ppm

I got 22.9119 for the first and 57080 for the next. i don't think sig figs are that important here.

1/26/2009 7:56:23 PM

For the first one, I'd suggest that you ponder the meaning of adjective "colligative." In other words, the 0.165 is m_c, or, put another way, m_c = 0.165 is the same as "colligative molality is 0.165m."

For the latter, you might note that % by wt would be wt fraction *100 = 0.1161 (or, put another way, 0.1161% = 0.001161). This sometimes causes decimal place errors which, if done by a pharmacist, can lead to interesting results.

Good luck.

[Edited on January 26, 2009 at 9:08 PM. Reason : add "]

1/26/2009 9:08:00 PM

^thank you, i figured it was something stupid for the second one. got that one right (2 decimal places off because i was in a hurry and messed up the %)

now off to figure out what colligative molality is

1/26/2009 9:21:23 PM

problem solved thank you tdub

1/26/2009 10:19:59 PM

more like thanks evan, amirite?

1/27/2009 10:39:35 AM

^for the first one. GreatGazoo caught my stupid error in the second one

1/27/2009 1:05:52 PM

All of the iron in 1.402 g of an iron sample is converted into Fe2+, which was then titrated with dichromate:
14H1+ + Cr2O72- + 6Fe2+ -> 6Fe3+ + 2Cr3+ + 7H2O
What is the mass percent iron in the sample if 31.77 mL of 0.09548 M K2Cr2O7 were required to reach the equivalence point?

can't even figure out where to begin and this is my last problem of the webassign

.007 moles of the 1.402 sample didn't get used, so .018 moles in the reaction/.027 supposed iron moles = 72% = the answer

1/28/2009 5:45:05 PM

My calibration curve is y= 1.44x+.0597

Calculate the molar absorptivity () of Cu2+ at 620 nm given that the cuvettes used in lab had a pathlength of 1.00 cm.

1/29/2009 10:24:04 PM

^1.44

that was made from the data you used in the lab.

the 1.44 represents the molar absorptivity at a wavelength times the pathlength (in cm)
so 1 cm * 1.44

1/29/2009 11:20:53 PM

HX is a weak acid that partially dissociates in water: HX -> H1+ + X1-. A 0.505-m aqueous solution of HX has a freezing point of -1.18 oC. Express your answers to the following to three significant figures.
b) (EXTRA CREDIT) What is the molality of X1- ions in the solution?

since it's extra credit i'm not going to ask for anyone to answer this before midnight (when my webassign is due) but could someone help me figure this out?

i know that i'd have to figure out how completely HX dissociates here, and I'm not sure how to determine that

1/30/2009 10:12:50 PM

english holmes, we're not all chemistry wizards.

jesus christ i'm glad all i had to take was CH101

I'm Big Business and i approved this message.

1/31/2009 1:01:25 AM

lol anyone wanna show me now that the webassign is past due? i know you're on here warren, pls to give further hint

1/31/2009 5:46:54 PM

^^ I just got thru 3/4 of an equation and realized a problem. I dont know how to calculate the molarity of ions but I found a tutorial and ill look at it tommorow and i can't use that fp in my calculation

this is retarded, ill look at it again in the mornign


Ego, how far did you get? What part in particular did you complete and what did you get stuck on

[Edited on February 1, 2009 at 1:03 AM. Reason : d]

2/1/2009 12:59:42 AM

Quote :

"HX is a weak acid that partially dissociates in water: HX -> H1+ + X1-. A 0.505-m aqueous solution of HX has a freezing point of -1.18 oC. Express your answers to the following to three significant figures. b) (EXTRA CREDIT) What is the molality of X1- ions in the solution?"

First you need to figure out i from the freezing point depression equation.

Freezing Point depression and molarity of the solute are related by:
?Tf = i · Kf · m

i - van't Hoff factor (unknown)
Kf - cryoscopic constant, which is 1.858 K·kg/mol for water
m - molality of the solute (0.505)
?Tf = 1.18


So 1.18 = i · 1.858 · 0.505 and i = 1.257.

Now calculate your percent dissociation. To do this, multiply the i you got in the previous equation (1.257) by 100/theoretical i. The theoretical i in this case is 2 because you have two ions on the right hand side of your dissociation equation.

theoretical i = 2
% dissociation = 1.257 x 100 / 2 = 62.85


To find the molality of X-, take your percent dissociation in percent form (0.6285) and multiply by the original molality (0.505)

molality of X- = 0.6285 * 0.505 m = 0.317

I hope that helps.

2/1/2009 1:25:23 AM

nobody in the chemistry world uses molality

2/1/2009 1:50:44 PM

I know right?

2/1/2009 4:10:11 PM

i didn't know if the van hoff factor had anything to do with it. i found that for part a, and got a 100 on everything else, but wasn't really sure how to figure out who completely the ion dissociated. thank you vix.

2/1/2009 4:30:10 PM

10/10 class today warren, 10/10. with the lols

2/5/2009 7:20:58 PM

this web assign wants me to calculate colligative molality. It gives me grams, the kf, and change in temp but no molar mass and no i. What am i doing wrong

2/5/2009 9:26:15 PM

You need to capitalize "i" and put some punctuation at the end of that sentence.

2/5/2009 9:33:30 PM

the variable "i" is not capitalized

2/5/2009 10:02:30 PM

^she was talking about when you said

Quote :

"What am i doing wrong"

2/5/2009 10:16:00 PM

At some temperature, K = 116 for the gas phase reaction
H2 + F2 2HF
What is the concentration of HF in an equilibrium mixture established by adding 2.88 mol each of H2 and F2 to a 1.00 L container at this temperature?

can anyone tell me where to start?

2/27/2009 6:09:22 PM

I'm guessing that your lecture notes from the last two days of class, that should have four worked examples of this sort of thing, are not an option?

[Edited on February 27, 2009 at 7:04 PM. Reason : ..]

2/27/2009 6:51:45 PM

hey greatgazoo someone told me you had tons of old tests and homework answers

let me have some please!!!!1111

2/27/2009 7:43:35 PM

^^i spilled water on my notes

2/27/2009 11:26:37 PM

the answer is 42.

3/5/2009 10:22:27 AM

Vix no need to hate because jen is more beautiful than you

I'm Big Business and i approved this message.

3/5/2009 2:46:01 PM

jen should come study with me

3/5/2009 2:48:42 PM

bro you're doin it wrong

I'm Big Business and i approved this message.

3/5/2009 3:06:18 PM

i'm not sending her penis pics

3/5/2009 5:11:48 PM

*sigh* all wrong bro. all wrong

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3/6/2009 12:48:11 AM

Determine the following in a 0.0770 M solution of HIO3 (Ka = 0.170)?

[HIO3] = ?

is it not .0770 M?

3/16/2009 9:44:43 PM

HIO3 + H20 -> IO3^1- + H3O^+

-Create reaction table, where you add X on both sides and subtract X from the reactants side. Thus, [HIO3] = .0770 - X, not just .0770.

3/17/2009 12:06:10 AM

^didn't think to do that, figured it'd want the initial conc. for whatever reason. oh well, got everything else right

3/17/2009 12:29:37 AM