Form 300 mL of 0.25 M Copper 2 sulfate pentahydrate

You want how many mols? That's M x L = mols.

Then mols = grams/molar mass. You know molar mass and mols, solve for grams.
Dissolve that many g CuSO4.5H2O in some water and make to exactly 300 mL with distilled water.

To prepare 300 mL of 0.25 M Copper (II) sulfate pentahydrate (CuSO4 · 5H2O) solution, you need to consider two things: the molar concentration and the molecular weight of the compound.

First, let's calculate the amount of Copper (II) sulfate pentahydrate needed using the formula:

Amount (in moles) = Concentration (in moles/L) × Volume (in liters)

Step 1: Convert the volume from milliliters (mL) to liters (L):
300 mL ÷ 1000 = 0.3 L

Step 2: Calculate the moles of Copper (II) sulfate pentahydrate:
Amount (in moles) = 0.25 mol/L × 0.3 L = 0.075 moles

The molar mass of Copper (II) sulfate pentahydrate is:
1 Copper (Cu) atom × atomic mass = 1 × 63.55 g/mol = 63.55 g/mol
1 Sulfur (S) atom × atomic mass = 1 × 32.07 g/mol = 32.07 g/mol
4 Oxygen (O) atoms × atomic mass = 4 × 16.00 g/mol = 64.00 g/mol
10 Hydrogen (H) atoms × atomic mass = 10 × 1.01 g/mol = 10.10 g/mol

The total molecular weight is:
63.55 + 32.07 + 64.00 + 10.10 = 169.72 g/mol

Step 3: Calculate the mass needed using the moles and molecular weight:
Mass = Amount (in moles) × Molecular weight
Mass = 0.075 moles × 169.72 g/mol
Mass ≈ 12.73 grams

Therefore, you need approximately 12.73 grams of Copper (II) sulfate pentahydrate to prepare 300 mL of 0.25 M solution. Dissolve this mass of the compound in water and adjust the volume to exactly 300 mL if needed.