# What would be the concentration of salt if 174*g NaCl were added to 1000*mL of water?

$\text{Concentration}$ $=$ $\left(\text{Moles of solute")/("Volume of solution}\right)$
If we have $1.00$ $k g$ of water, there is $1.00$ $L$ of water. (Note that volume would not change substantially upon dissolution of the salt).
Thus, $\text{Molarity}$ $=$ $\frac{\frac{174 \cdot \cancel{g}}{58.44 \cdot \cancel{g} \cdot m o {l}^{-} 1}}{1.00 \cdot L}$ $\approx$ $3$ $m o l \cdot {L}^{-} 1$