# 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?

Jul 26, 2017

Well.......$2.25 \cdot g$ solute are needed.

#### Explanation:

By definition, $\text{concentration"="moles of solute"/"volume of solution}$.........

And also $\text{moles of solute"="mass of solute"/"molar mass of solute}$......

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

$500 \times {10}^{-} 3 \cdot L \times 0.077 \cdot m o l \cdot {L}^{-} 1 \times 58.44 \cdot g \cdot m o {l}^{-} 1 = 2.25 \cdot g$

The $\text{osmolarity}$ of the solution is equal to $\text{2"xx"molarity}$ $=$ $2 \times 0.077 \cdot m o l \cdot {L}^{-} 1$ because there are 2 equiv of ions in solution.....