noelpj
Harmless
Posts: 8
Registered: 19-9-2004
Member Is Offline
Mood: No Mood
|
|
help with homework
Hi I need help with some homework:
1. a 10.00 g sample of CaCl2(Calcium Chloride) is added to water to make 100.0 mL of solution. Then a 400.0 mL
sample of water is added to this solution. Determine the concentration of Cl-(negative) ions in the dilluted solution.
2. a 50 mL sample sample of 0.85 mol/L NaHCO3 is diluted to a volume of 250.0 mL. Then a 50.0 mL sample of
this dilute solution is evaporated to dryness. What mass of NaHCO3 remains?
3. A 13.6 g sample of NaCl and a 7.34 g sample of CaCl2 are dissolved in water to make 200 mL of solution.
What is the concentration of Cl- in this solution?
4. A 50 g sample of Al(No3)3 is dissolved in water to prepare 1500 mL of
solution. What is the concentration, in mol/L of No3- ions in the solution?
The sooner the better
|
|
|
Quantum
Hazard to Others
Posts: 300
Registered: 2-12-2003
Location: Nowhereville
Member Is Offline
Mood: Interested
|
|
Those are all basic math questions. Any good textbook will show you how to do them. Did you fall asleep that day in class?
What if, what is isn\'t true?
|
|
|
BromicAcid
International Hazard
Posts: 3245
Registered: 13-7-2003
Location: Wisconsin
Member Is Offline
Mood: Rock n' Roll
|
|
| Quote: |
1. a 10.00 g sample of CaCl2(Calcium Chloride) is added to water to make 100.0 mL of solution. Then a 400.0 mL sample of water is added to this
solution. Determine the concentration of Cl-(negative) ions in the dilluted solution.
|
Concentration in... parts per billion, parts per million, molarity, percent?
| Quote: | | 2. a 50 mL sample sample of 0.85 mol/L NaHCO3 is diluted to a volume of 250.0 mL. Then a 50.0 mL sample of this dilute solution is evaporated to
dryness. What mass of NaHCO3 remains? |
Determine the molarity of the final solution after dilution using the normal formula:
(Mi)x(Vi) = (Mf)x(Vf)
Then after you have the molarity you can figure out how many mols of NaHCO3 you have in your sample and therefor how many mols you have upon drying.
| Quote: | | 3. A 13.6 g sample of NaCl and a 7.34 g sample of CaCl2 are dissolved in water to make 200 mL of solution. What is the concentration of Cl- in this
solution? |
Again, what units?
| Quote: | | 4. A 50 g sample of Al(No3)3 is dissolved in water to prepare 1500 mL of solution. What is the concentration, in mol/L of No3- ions in the solution?
|
1500 ml is 1.5 L, how many mol's of Al(NO3)3 are there? For each mol of Al(NO3)3 in 50 g there are 3 mols NO3- now you have your mols NO3- and
your volume, divide.
These are simple problems and the semester is just starting, if you don't master these basics you will not pass the class, you can post any
answers you figure out here and we can check them for you, be sure to include pertinent calculations you come up with so if you have a mistake
somewhere we can spot it and tell you how to correct it in the future.
|
|
|
noelpj
Harmless
Posts: 8
Registered: 19-9-2004
Member Is Offline
Mood: No Mood
|
|
| Quote: | Originally posted by Quantum
Those are all basic math questions. Any good textbook will show you how to do them. Did you fall asleep that day in class? |
actually its review from last year and text books generalize when they teach
|
|
|
noelpj
Harmless
Posts: 8
Registered: 19-9-2004
Member Is Offline
Mood: No Mood
|
|
| Quote: | Originally posted by BromicAcid
| Quote: |
1. a 10.00 g sample of CaCl2(Calcium Chloride) is added to water to make 100.0 mL of solution. Then a 400.0 mL sample of water is added to this
solution. Determine the concentration of Cl-(negative) ions in the dilluted solution.
|
Concentration in... parts per billion, parts per million, molarity, percent?
| Quote: | | 2. a 50 mL sample sample of 0.85 mol/L NaHCO3 is diluted to a volume of 250.0 mL. Then a 50.0 mL sample of this dilute solution is evaporated to
dryness. What mass of NaHCO3 remains? |
Determine the molarity of the final solution after dilution using the normal formula:
(Mi)x(Vi) = (Mf)x(Vf)
Then after you have the molarity you can figure out how many mols of NaHCO3 you have in your sample and therefor how many mols you have upon drying.
| Quote: | | 3. A 13.6 g sample of NaCl and a 7.34 g sample of CaCl2 are dissolved in water to make 200 mL of solution. What is the concentration of Cl- in this
solution? |
Again, what units?
| Quote: | | 4. A 50 g sample of Al(No3)3 is dissolved in water to prepare 1500 mL of solution. What is the concentration, in mol/L of No3- ions in the solution?
|
1500 ml is 1.5 L, how many mol's of Al(NO3)3 are there? For each mol of Al(NO3)3 in 50 g there are 3 mols NO3- now you have your mols NO3- and
your volume, divide.
These are simple problems and the semester is just starting, if you don't master these basics you will not pass the class, you can post any
answers you figure out here and we can check them for you, be sure to include pertinent calculations you come up with so if you have a mistake
somewhere we can spot it and tell you how to correct it in the future. |
For number 1 and number 3 the units is molarity. For number 2, how did you end up with moles, your answer is mass which should be grams(g). For the
last question you find the moles by dividing mass over molar mass.
|
|
|
BromicAcid
International Hazard
Posts: 3245
Registered: 13-7-2003
Location: Wisconsin
Member Is Offline
Mood: Rock n' Roll
|
|
Then for numbers one and three turn the grams of solutes into mols.
For example, in number one the final volume is .5L and the mols would be the grams of calcium chloride divided by the grams per mol. You have your
mols of CaCl2 and divide by .5 now you have your mols of CaCl2 in molarity, for each mole of CaCl2 how many mols of Cl2 are there? Hint: There are two Then multipy by that number.
Number three is essntially the same, just do the CaCl2 and the NaCl seperately and add them together.
Number two I instructed you to get all the way to the point to determine the number of mols left on drying, just turn this into grams again and
you're set.
And for the last question my question to you was a hint on how to finish the problem, I gave you pretty much everything you need to solve it.
A periodic table is all that is needed, there are plenty free online.
|
|
|
noelpj
Harmless
Posts: 8
Registered: 19-9-2004
Member Is Offline
Mood: No Mood
|
|
Thanks for the help. I'm only having one more problem:
I'm having trouble with number 2 still. Here is what I did step by step:
1. Found Final Concentration(molarity) using C2=C1*V1/V2
2. Multiplied final concentration by .05 L to find moles.
3. Multiplied moles by molar mass to find mass
The answer should be 0.71g but I'm getting 0.071g
|
|
|
BromicAcid
International Hazard
Posts: 3245
Registered: 13-7-2003
Location: Wisconsin
Member Is Offline
Mood: Rock n' Roll
|
|
| Quote: | | 2. a 50 mL sample sample of 0.85 mol/L NaHCO3 is diluted to a volume of 250.0 mL. Then a 50.0 mL sample of this dilute solution is evaporated to
dryness. What mass of NaHCO3 remains? |
(50 ml) x (.85M) = (250 ml) x (Mf)
42.5 = (250 ml) x (Mf)
0.17 = Mf
0.17 x .05 = .0085 Mols
.0085 Mols x (84 g/mol) = 0.714 g
Maybe you lost a decimal place somewhere?
|
|
|
noelpj
Harmless
Posts: 8
Registered: 19-9-2004
Member Is Offline
Mood: No Mood
|
|
why didnt you convert to L?
|
|
|
BromicAcid
International Hazard
Posts: 3245
Registered: 13-7-2003
Location: Wisconsin
Member Is Offline
Mood: Rock n' Roll
|
|
Because the units cancel when converting the molarity initial to the molarity final. So the units don't matter.
With liters or with ml it should turn out the same.
(50 ml) x (.85M) = (250 ml) x (Mf)
42.5 = (250 ml) x (Mf)
0.17 = Mf
(.050 L) x (.85M) = (0.250 L) x (Mf)
.0425 = (0.250 L) x (Mf)
0.17 = Mf
[Edited on 9/20/2004 by BromicAcid]
|
|
|
noelpj
Harmless
Posts: 8
Registered: 19-9-2004
Member Is Offline
Mood: No Mood
|
|
Thanks alot for the help. I know I will be coming to this site next time I need more help. Thanks again
|
|
|
![[*]](images/xpblue/default_icon.gif){kind=icon}
