How do I solve this?

How many grams of solid NaOH are required to prepare 200 mL of a 0.05 M solution? What will be the concentration expressed in % W/V?

For the first part, just remember the definition. molarity = # mols/Liter of solution.

AND mols = grams/molar mass.
Using equation 1, you know molarity and you know liters, calculate mols.
Using equation 2, you know mols and molar mass, calculate grams.

To solve this problem, we need to use the formula relating molarity, volume, and molar mass.

Step 1: Calculate the number of moles of NaOH needed.
The formula for calculating the number of moles is:
moles = molarity x volume (in L)

Given:
Molarity (M) = 0.05 M
Volume (in L) = 200 mL = 200/1000 = 0.2 L

moles = 0.05 M x 0.2 L = 0.01 moles

Step 2: Convert moles to grams.
To convert moles to grams, we will need the molar mass of NaOH. The molar mass of sodium (Na) is 22.99 g/mol, the molar mass of oxygen (O) is 16.00 g/mol, and the molar mass of hydrogen (H) is 1.01 g/mol. Adding these up, we get:

Molar mass of NaOH = 22.99 g/mol + 16.00 g/mol + 1.01 g/mol = 40.00 g/mol

grams = moles x molar mass = 0.01 moles x 40.00 g/mol = 0.4 grams

Thus, 0.4 grams of solid NaOH are required to prepare 200 mL of a 0.05 M solution.

Step 3: Calculate the concentration expressed in % W/V.
The formula for calculating % W/V is:

% W/V = (mass of solute / volume of solution) x 100

Given:
Mass of solute (NaOH) = 0.4 grams
Volume of solution = 200 mL

% W/V = (0.4 grams / 200 mL) x 100 = 0.2 g/mL x 100 = 0.2%

Therefore, the concentration expressed in % W/V is 0.2%.