# What is the concentration if .358 g of NaNO_3 is present in 555 mL of water?

$7.59 \times {10}^{- 3} m o l \cdot {L}^{-} 1$ in sodium nitrate
So $\frac{0.358 \cdot \cancel{g}}{85.00 \cdot \cancel{g} \cdot m o {l}^{-} 1}$ $\times$ $\frac{1}{555 \times {10}^{- 3} \cdot L}$ $=$ ??*mol*L^-1