Well, assuming you have an aqueous solution of #NaCl#, there exist free #Na^+# and #Cl^-# ions. The equation for that is:
#NaCl(aq)->Na^+(aq)+Cl^(-)(aq)#
So, there is a #1:1# ratio between sodium cations and chlorine anions.
In a #1 \ "M"# sodium chloride solution, there exist #1 \ "mol/L"# of sodium chloride.
And so, we get,
#(1 \ "mol")/(color(red)cancelcolor(black)"L")*(1color(red)cancelcolor(black)"L")/(1000color(red)cancelcolor(black)"mL")*100color(red)cancelcolor(black)"mL"=0.1 \ "mol"#
In a mole of sodium ions, there exist #6.02*10^23# sodium ions. So here, there are a total of:
#0.1color(red)cancelcolor(black)("mol of" \ Na^+)*(6.02*10^23 \ Na^+ \ "ions")/(color(red)cancelcolor(black)("mol of" \ Na^+))=6.02*10^22 \ Na^+ \ "ions"#
