how many grams of solid NaOH are required to prepare 200.0 mL of a .050 M solution?

mL x M = mL x M

200/1000=0.2

0.2*0.05=0.01
0.01*40=0.4g

To determine the number of grams of solid NaOH required to prepare a 0.050 M solution, you need to use the molarity formula and convert the volume to the number of moles. Here's how you can calculate it:

Step 1: Convert the given volume from milliliters (mL) to liters (L):
200.0 mL = 200.0/1000 = 0.200 L

Step 2: Use the molarity formula: Molarity (M) = Moles (mol) / Volume (L).

0.050 M = Moles (mol) / 0.200 L

Step 3: Rearrange the formula to solve for Moles (mol):
Moles (mol) = Molarity (M) x Volume (L)

Moles (mol) = 0.050 M x 0.200 L

Moles (mol) = 0.010 mol

Step 4: Use the molar mass of NaOH to convert the moles to grams. The molar mass of NaOH is approximately 40.00 g/mol (22.99 g/mol for Na + 16.00 g/mol for O + 1.01 g/mol for H).

Grams = Moles (mol) x Molar mass (g/mol)

Grams = 0.010 mol x 40.00 g/mol

Grams = 0.40 g

Therefore, you will need approximately 0.40 grams of solid NaOH to prepare 200.0 mL of a 0.050 M solution.