Answer
There are necessary 480. g of $NaOH$ to prepare that solution
Work Step by Step
1. Since the molarity of that solution is equal to $ 6.00 \space M \space NaOH$:
$1 \space L \space solution = 6.00 \space moles \space NaOH$
$\frac{1 \space L \space solution}{ 6.00 \space moles \space NaOH} $ and $\frac{ 6.00 \space moles \space NaOH}{1 \space L \space solution}$
2. Determine the molar mass of this compound (NaOH), and setup the conversion factors:
Molar mass :
$Na: 22.99g $
$O: 16.00g $
$H: 1.008g$
22.99g + 16.00g + 1.008g = 40.00g
$ \frac{1 \space mole \space (NaOH)}{ 40.00 \space g \space (NaOH)}$ and $ \frac{ 40.00 \space g \space (NaOH)}{1 \space mole \space (NaOH)}$
3. Use the conversion factors to calculate the mass of solute in $ 2.00$ L of that solution:
$ 2.00 \space L \space solution \times \frac{ 6.00 \space moles \space NaOH}{1 \space L \space solution} \times \frac{ 40.00 \space g \space NaOH}{1 \space mole \space NaOH} = 480. \space g \space NaOH$
