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

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Answer 1 · 2017-07-26T17:40:32

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

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.....

How do we prepare a 500*mL500*mL volume of 0.077*mol*L^-10.077*mol*L^-1 solution with respect to sodium chloride? What is the "osmolarity""osmolarity" of this solution?