# Question #11ea0

Approximately $20.5 g$ of $N a C l$ are in the $0.350 M$ aqueous solution.
$M M \left[N a C l\right] = \frac{58.5 g}{m o l}$
$\frac{0.350 m o l}{L} \cdot \frac{58.5 g}{m o l} \approx 20.5 g$