questions of dilution & gravimetric
 

I am now studying on my test and I came upon one question which I can't answer but I still remember that the answer is 1.09 M. Yeah, I have a bad memory... So, here goes:

A 46.2-mL, o.568 M calcium Nitrate [Ca(NO3)2] solution i smixed with 80.5 mL of 1.396 M calcium nitrate solution. What is the concentration? :shock:

ALSO....

Is it true that according to my teacher that we must use distilled water in the gravimetric analysis of chlorides?? Why is it so? Isn't it just the same if we use regular drinking water?

And yesterday, I have to answer this super tough question is the school and I got zero since I am not really good in chemistry and I'm a bit too shy to ask anything...This is the question:

If 30.0 mL of 0.150 M CaCl2 is added to 15.0 mL of 0.100 M AgNO3, what is the mass in grams of AgCl precipitate?

Thanks so much for helping me.
:oops:

Re: questions of dilution & gravimetric
 

use c = n/V to calculate the amount of mmol Ca(NO3)2 of the 46.2 mL 0.568 M solution and the same way for the 80.5 mL 1.396 M Ca(NO3)2 solution.

Add the mols and volumes and use again c = n/V to calculate the final concentration Ca(NO3)2. The answer will be 1.09 M

First: what is the balanced reaction equation ?

Then you need to calculate the amount of mol CaCl2 and AgNO3 using c = n/V.

Then determin which is the limiting reagent and use stoichiometry to calculate the amount of mol AgCl formed.

--> convert amount of mol AgCl to g. The answer will be 0.21 g.

hey are you sure the answer is 0,21 g?? I tried for about 10 times but asnwer is much bigger than that. :(

i'm 100% sure about that. 1,5 mmol AgCl will be formed and thus 215 mg.

But we need to balance the equation right?
So it's like this:
2 AgNO3 + CaCl2 --> 2 AgCl + Ca(NO3)2
1,5 mole 4,5 mole

then.... we divide the 1,5 mole with 2 = 0,75 mole
and then finding the mass of AgCl = 2/2 x 0,75 x 143,35 = 107, 5 mg.

AgNO3 and AgCl are in equimolar equivalents of 1.5 mmol

1.5 x 10(-3) mol x 143.5 g/mol = 0.21 g