What the "millimolarities"millimolarities with respect to sodium and chloride ions, of a 9.00*g9.00⋅g mass of sodium chloride dissolved in a volume of 1*L1⋅L of solution...?

1 Answer
anor277
Jun 25, 2017

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

And here, we have a concentration of 154*mmol*L^-1154⋅mmol⋅L−1.

Explanation:

"Molarity NaCl"=((9.00*g)/(58.44*g*mol^-1))/(1.000*L)=0.154*mol*L^-1Molarity NaCl=9.00⋅g58.44⋅g⋅mol−11.000⋅L=0.154⋅mol⋅L−1

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

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

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

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