Answer
There are necessary 120. g of $NaOH$ to prepare that solution.
Work Step by Step
1. Since the molarity of that solution is equal to $1.50 \space M \space NaOH$:
$1 \space L = 1.50 \space moles \space NaOH$
$\frac{1 \space L}{ 1.50 \space moles \space NaOH} $ and $\frac{ 1.50 \space moles \space NaOH}{1 \space L}$
2. Calculate the molar mass of $NaOH$:
$Na: 22.99g $
$O: 16.00g $
$H: 1.008g$
22.99 g + 16.00 g + 1.008 g = 40.00 g
$ \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 in 2.00 L of that solution:
$2.00 \space L \times \frac{ 1.50 \space moles \space NaOH}{1 \space L} \times \frac{ 40.00 \space g \space (NaOH)}{1 \space mole \space (NaOH)} = 120. \space g \space NaOH$
