How many grams of CuSO4 * 5H2O are needed to prepare 100mL of a 0.10 M solution ?

To determine the number of grams of CuSO4 * 5H2O needed to prepare a 0.10 M solution, you will need to follow these steps:

Step 1: Determine the molar mass of CuSO4 * 5H2O.
- CuSO4: Copper (Cu) has a molar mass of 63.55 g/mol. Sulfur (S) has a molar mass of 32.07 g/mol, while each of the four oxygen atoms (O) has a molar mass of 16.00 g/mol. Therefore, the molar mass of CuSO4 is:
(63.55 g/mol) + (32.07 g/mol) + (4 × 16.00 g/mol) = 159.61 g/mol.
- 5H2O: Each water molecule (H2O) has a molar mass of (2 × 1.01 g/mol) + (16.00 g/mol) = 18.02 g/mol. Since there are five water molecules in CuSO4 * 5H2O, the molar mass is: 5 × 18.02 g/mol = 90.10 g/mol.
- Adding these two molar masses together gives: 159.61 g/mol + 90.10 g/mol = 249.71 g/mol.

Step 2: Calculate the number of moles needed for the desired concentration.
- The amount of moles (n) can be calculated using the equation: n = M × V, where M is the molarity (mol/L) and V is the volume (L).
- We have the desired molarity (0.10 M) and volume (100 mL = 0.100 L).
- Plugging in these values into the equation: n = 0.10 mol/L × 0.100 L = 0.010 mol.

Step 3: Convert moles to grams.
- To convert moles to grams, we need to multiply the number of moles by the molar mass.
- We already calculated the molar mass of CuSO4 * 5H2O in Step 1 (249.71 g/mol).
- Multiplying the molar mass by the number of moles gives: 249.71 g/mol × 0.010 mol = 2.497 g.

Therefore, you would need approximately 2.497 grams of CuSO4 * 5H2O to prepare 100 mL of a 0.10 M solution.

mass=molarity*molemass*volumeinLiters

in determining mole mass, include the water of hydration (5H2O)

O.1