Calculating Weight of Unknown Sample for 0.02M Cl- Solution

[higherme]

I'm not sure if I am doing this correctly... can someone check for me please.. thanks!

Calculate the weight of the unknown sample needed to produce a 40mL [Cl-] solution of about 0.02M. Assume that the weight percent of NaCl in the unknown sample is 50%.

My answer:


(0.02 mol/L) x (0.04L) x (22.990g NaCl/ mol)
= 0.018392 g x 2 <=== because the sample contains only 50% NaCl
= 0.036784 g

is that the right way? Thanks again.

and does anyone know why it has to be 0.02M? is it because we are going to use 0.02M of AgNO3 to precipitate out the Cl-

 

[at3rg0]

Correct me if I'm wrong, but aren't there more than 22.990g NaCl/mole?

 

[higherme]

ooops

the molar mass of NaCl should be 58.443g/mol

so it would be:
(0.02 mol/L) x (0.04L) x (58.443g NaCl/ mol)
=0.0467544g x 2
0.0935 g

 

[higherme]

is it right now? :P

 

[higherme]

and does anyone know why it has to be 0.02M [Cl-]? is it because in the experiment, we are going to use 0.02M of AgNO3 to precipitate out the Cl-??

 

[eli64]

that would seem reasonable, then it would be a 1:1 volume ratio you would need of AgNO3