How do we prepare a #500*mL# volume of #0.077*mol*L^-1# solution with respect to sodium chloride? What is the #"osmolarity"# of this solution?

1 Answer
Jul 26, 2017

Answer:

Well.......#2.25*g# solute are needed.

Explanation:

By definition, #"concentration"="moles of solute"/"volume of solution"#.........

And also #"moles of solute"="mass of solute"/"molar mass of solute"#......

And thus we need...................

#500xx10^-3*Lxx0.077*mol*L^-1xx58.44*g*mol^-1=2.25*g#

The #"osmolarity"# of the solution is equal to #"2"xx"molarity"# #=# #2xx0.077*mol*L^-1# because there are 2 equiv of ions in solution.....