# If you have exactly 1.00 L of sodium chloride solution that masses 1050 g, what is the molarity of the solution?

May 30, 2016

Approx. $1.4 \cdot m o l \cdot {L}^{-} 1$.
There are $\left(1050 - 1000\right) \cdot g$ of solute in the AQUEOUS solution:
$\text{Concentration}$ $=$ $\text{Moles of solute"/"Volume of Solution}$
$=$ $\frac{50 \cdot g}{36.46 \cdot g \cdot m o {l}^{-} 1} \times \frac{1}{1.00 \cdot L}$ $=$ ??mol*L^-1