Question

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...?

Answer 1

`"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^-]??`