Chemistry: Molecular Approach (4th Edition)

37 g of $NaNO_3$ are needed.
$NaNO_3$ : ( 22.99 $\times$ 1 )+ ( 14.01 $\times$ 1 )+ ( 16.00 $\times$ 3 )= 85.00 g/mol - Use all the information as conversion factors: $$400.0 \space mL \times \frac{1 \space L}{1000 \space mL} \times \frac{ 1.1 \space mol}{1 \space L} \times \frac{ 85.00 \space g}{1 \space mol} = 37 \space g$$