What the #"millimolarities"# with respect to sodium and chloride ions, of a #9.00*g# mass of sodium chloride dissolved in a volume of #1*L# of solution...?

1 Answer
Jun 25, 2017

Answer:

#"Molarity"="Moles of solute"/"Volume of solution"#.

And here, we have a concentration of #154*mmol*L^-1#.

Explanation:

#"Molarity NaCl"=((9.00*g)/(58.44*g*mol^-1))/(1.000*L)=0.154*mol*L^-1#

But we were asked to supply #"millimolarity"# of #Na^(+)(aq)#.

Because sodium chloride reacts in solution to give equimolar #Na^+#, and #Cl^-# as the aquated ions....

#[Na^+]=[NaCl(aq)]=0.154*mol*L^-1#

And..........

#[Na^+]=0.154*mol*L^-1=0.154*cancel(mol)*L^-1xx10^3*mmol*cancel(mol^-1)=154*mmol*L^-1.#

And what is #[Cl^-]??#