Answer:
120 grams of NaOH
Step-by-step explanation:
This can be solved by using Unit Conversion.
Let n = # grams of NaOH needed to prepare 2.0 L of a 1.5 M NaOH solution.
Known variables
V = 2.0 L
[NaOH] = 1.5 mol/L
Molecular Weight of NaOH = 39.997 g/mol
n = 2.0 L x (1.5 mol/L) x (39.997 g NaOH/ mol) = 119.991 grams
rounding to two significant figures
n = 120 grams of NaOH