i find it confusing on how to do this question..

what mass of solute is needed to prepare 400 mL of 0.850 M CuSO4 from CuSO4.5H2O

Molar mass CuSO4.5H2O = 249.69 g/mol

You want 400ml of 0.85M: 249.69 *400/1000*0.85 = 84.9g CuSO4.5H2O dissolved to total volume 400ml.

The confusing part here is that you want a solution of CusO4 but you have given CuSO4.5H2O to make it.

What to remember. 1 mol CuSO4 = 1 mol CuSO4.5H2O; therefore, you want how many mols CuSO4? That's
M x L = 0.850 x 0.400 = ?
But that's the same number of mols you want for CuSO4.5H2O also; therefore
g = mols x molar mass and
g CuSO4.5H2O = mols CuSO4 x molar mass CuSO4.5H2O

To solve this question, you need to use the formula:

\( M_1V_1 = M_2V_2 \)

Where:
- \( M_1 \) is the initial molarity of the solution (CuSO4.5H2O)
- \( V_1 \) is the initial volume of the solution (unknown)
- \( M_2 \) is the final molarity of the solution (0.850 M CuSO4)
- \( V_2 \) is the final volume of the solution (400 mL)

First, let's find the initial volume (\( V_1 \)) of the CuSO4.5H2O solution by rearranging the equation:

\( V_1 = \frac{{M_2V_2}}{{M_1}} \)

Now, let's substitute the known values into the equation:

\( V_1 = \frac{{0.850 \, \text{M} \times 400 \, \text{mL}}}{{\text{molarity of CuSO4.5H2O}}} \)

To find the molarity of CuSO4.5H2O, we need to consider its molecular weight. The molecular weight of CuSO4 is 159.62 g/mol, and the molecular weight of H2O is 18.02 g/mol. Since there are 5 water molecules (H2O) in each formula of CuSO4.5H2O, the total molecular weight is:

\( \text{molecular weight of CuSO4.5H2O} = \text{formula weight of CuSO4} + 5 \times \text{molecular weight of H2O} \)

\( \text{molecular weight of CuSO4.5H2O} = 159.62 \, \text{g/mol} + 5 \times 18.02 \, \text{g/mol} \)

Now that we have the molecular weight of CuSO4.5H2O, substitute it back into the equation to solve for \( V_1 \):

\( V_1 = \frac{{0.850 \, \text{M} \times 400 \, \text{mL}}}{{\text{molecular weight of CuSO4.5H2O}}} \)

Finally, to find the mass of solute (CuSO4.5H2O) needed, multiply the obtained \( V_1 \) by the molecular weight of CuSO4.5H2O:

\( \text{mass of solute} = \text{molecular weight of CuSO4.5H2O} \times V_1 \)