Answer
There are necessary 44.6 g of $NaNO_3$ to prepare that solution.
Work Step by Step
1. Since the molarity of that solution is equal to $ 3.00 \space M \space NaNO_3$:
$1 \space L \space solution = 3.00 \space moles \space NaNO_3$
$\frac{1 \space L \space solution}{ 3.00 \space moles \space NaNO_3} $ and $\frac{ 3.00 \space moles \space NaNO_3}{1 \space L \space solution}$
2. Determine the molar mass of this compound (NaNO_3), and setup the conversion factors:
Molar mass :
$Na: 22.99g $
$N: 14.01g $
$O: 16.00g * 3= 48.00g $
22.99g + 14.01g + 48.00g = 85.00g
$ \frac{1 \space mole \space (NaNO_3)}{ 85.00 \space g \space (NaNO_3)}$ and $ \frac{ 85.00 \space g \space (NaNO_3)}{1 \space mole \space (NaNO_3)}$
3. To convert from milliliters to liters: 1 L = 1000 mL
$\frac{1 \space L}{1000 \space mL}$ and $\frac{1000 \space mL}{1 \space L}$
3. Use the conversion factors to calculate the mass of solute in $ 175$ mL of that solution:
$ 175 \space mL \space solution \times \frac{1 \space L}{1000 \space mL} \times \frac{ 3.00 \space moles \space NaNO_3}{1 \space L \space solution} \times \frac{ 85.00 \space g \space NaNO_3}{1 \space mole \space NaNO_3} = 44.6 \space g \space NaNO_3$
