## a few more questions for the masters

**Anonymous**

the following questions are really confusing me and i have no idea one which route to take on them. Any generall assistance with them would be appreciated.

- A hospital saline solution is analyzed to confirm its concetration. A 50.0ml samle with a mass of 50.320g is evaporated to dryness. If the solid sodium chloride residue has a mass of .453g, find the molar concertraion of the NaCL solution.

- What are the respective concertrations (M) of FE^3+ and I^- afforded by dissolving .200 mol FeI3 in water and diulting to 725 mL?

- a 230mL sample of a .276 M CaCl2 solution was left on a hot plate overnight; the following moring , the solution is 1.10M. What colume of wate evaporated from the .276M CaCl2 solution?

- Predict the products of the folliwing reactions; write the balanced reaction and the net ionic equation for the reaction:

1) K3PO4(aq) + Ca(NO3)2(aq) -->

Anonymous| 17 July 2005 - 4:42pmim not eve nasking for answers here im just asking what type of problems these are...like for instance what equations i should be using becuase im trying to look for examples in my text book and there dont seem to be anything quite like these.

Anonymous| 17 July 2005 - 6:52pm## Re: a few more questions for the masters

These are classic stoichiometry questions. Just do a lot of those until you get the idea. There are many formulas for problems like these but the main idea is just playing around with units. The main equation for molarity is: c=n/V

I'll start with 2. which is easiest.

You have the reaction FeI3(s)->Fe3+ (aq) + 3 I- (aq)

When you start with .200 mol FeI3 you get .200 mol Fe3+.

In 725mL you have the Fe3+ concentration .200mol/.725L=.276mol/L (c=n/V)

Out of every FeI3 unit you get 3 I-.

So your iodine conc. is: .600mol/.725L=.828mol/L

3. is just as simple:

for example go by the amount of substance (n=cV)

n=.230L*.276mol/L=.06348mol

the new volume is:

V=n/c=.06348mol/1.10(mol/L)=.0577L

(you subtract that amount from the original)

I think in 1. you don't need the mass of the solution at all.

First you calculate the molar mass (M) of NaCl

M=(23+35.5)g/mol

then this is divided through the mass of NaCl (m=.453g): n=m/M

then you go: c=n/V

Anonymous| 17 July 2005 - 7:27pmright before i saw your post i hadthis sudden epiphany as to how to solve the problem...but with your help i checked it and i worked it out correctly, thank you so much! the answer i got for #1 was .155 M because i subtracted .

Thank you so damn much!