[Marcus Soutlo]

How do I find [NaI] from the following information given?

4.0 mL volume of 1.0 M NaI

I assume I need to use the formula n = c /v

Am I correct in this manner?

[Arkcon]

Depends on what you're really asking, what is the definition of [NaI]?  What do the brackets mean?

[Marcus Soutlo]

The brackets, dont they represent concentration, or more specifically, concentration of NaI?

[Arkcon]

Great, that's the definition I was taught, as well.  So, you want the concentration, presumably in Molarity?, of NaI.  That's a given in this question, so that's done.  Did you mix up questions?  Not to be unnecessarily critical, but you, Frederick95, et al tend to do this a lot, change questions mid-thread, and leave us to debate our own postings, with ourselves.

Does it mean moles?  If so, if you're given molarity and volume, can you compute moles?  Generally, n, means moles, for at least some formulas.  You appear to have memorized a separate formula for n, I think you'd be better off understanding the Molarity=moles / L, and rearrange that one, whenever you're given any two, and asked for the third.  You'll have enough to do, with trying to determine what you've really been given, and what you're really asked, as in this case.

[Marcus Soutlo]

Read my question entitled "order of reaction" by Marcus Soulto. Dont I have to use volume in my formula?

[Marcus Soutlo]

This website says that n = c / v is Molarity though?

http://www.chembuddy.com/?left=concentration&right=molarity

[Marcus Soutlo]

If i am given volume (4 mL), why not use this forumla?

Cant I find number of moles by dividing 1.0 M by the molar mass of NaI

then use c = n / v to find concentration?

Thanks again

[Arkcon]

Did you mix up questions?  Not to be unnecessarily critical, but you, Frederick95, et al tend to do this a lot, change questions mid-thread, and leave us to debate our own postings, with ourselves.

 

Read my question entitled "order of reaction" by Marcus Soulto. Dont I have to use volume in my formula?

 

If i am given volume (4 mL), why not use this forumla?

Cant I find number of moles by dividing 1.0 M by the molar mass of NaI

No, that's not how that's done.

then use c = n / v to find concentration?

Thanks again

[Marcus Soutlo]

Hmmmm.. how about

since I have 4 mL of 1 molar NaI..I multiply 1 X .004 and I get the concentration..which should be .004

[Marcus Soutlo]

"So, you want the concentration, presumably in Molarity?, of NaI.  That's a given in this question, so that's done."

How about finding the concentration of H2O2 given the following information?

Volume of 6% H2O2 (mL):10

[Borek]

Man, start thinking.

IF IT IS 1 MOLAR IT MEANS ITS CONCENTRATION IS 1 MOL/LITER.

What you are probably looking for is a way of calculating NaI concentration once it is diluted in your reaction mixture. I will now post formula that you have to use, then we will spend next two days debating this formula and trying to find out how to solve it for new concentration, finally you will ask some completely new and unrelated question in separate thread.

C₁V₁ = C₂V₂

Where C₁ and C₂ are concentrations before and after dilution and V₁ and V₂ are volumes before and after.

And I have thought I am patient 

But you have posted next question even before I have finished writing...

How about finding the concentration of H2O2 given the following information?

Volume of 6% H2O2 (mL):10

I believe you were already told how to do that as well. Find mass of solute, convert it to number of moles, than C=n/V.

[Borek]

This website says that n = c / v is Molarity though?

http://www.chembuddy.com/?left=concentration&right=molarity

No, it doesn't say that.

[Marcus Soutlo]

ummmm using calculation i have determined that consentration is 2.04 molar

what i did was: .06*.004L then i multiply this answer time 34 g (because that is molecular weight of hyrodgen peroxide) then i divide that by .004 LOL and i get 2.04

Using  C=n/V gives me 29 M which is too much

[Borek]

we will spend next two days (...) trying to find out how to solve it for new concentration

Bingo.

http://www.chemicalforums.com/index.php?topic=24524