How many grams of KOH are needed to make 350 mL of 2.0 M solution?

Calculations on solution would be based on the following mathematical formulae:
[1] No. of moles = Molarity × (Volume in L)
[2] No. of moles = (Mass of solute) / (Molar mass)

You are given the molarity (2.0 M) and the volume (350/1000 L = 0.35 L) of the solution.
According to [1]:
No. of moles of KOH = Molarity × Volume = (2.0 mol/L) × (0.35 L) = 0.7 mol

Molar mass of KOH = (39.0 + 16.0 + 1.0) g/mol = 56.0 g/mol

Now, (Molar mass) and (No. of moles) are known. Rearrangement [2]:
Mass of KOH = (No. of moles) × (Molar mass) = (0.7 mol) × (56.0 g/mol) = 39.2 g

2.0 M KOH means one liter of the solution contains two moles of potassium hydroxide. To determine the number of moles, multiply this number by the volume in liters. To determine the mass of potassium hydroxide, multiply the number of moles by the mass of one mole.

n = 0.350 * 2 = 0.7
Mass of one mole = 39.1 + 16 + 1 = 56.1 grams
Mass = 0.7 * 56.1 = 39.27 grams

If you round to two significant digits the answer will be 39 grams. I hope this is helpful for you.