In an experiment designed to determine the concentration of Cu+2 ions in an unknown solution, you need to prepare 100 mL of 0.10 M CuSO4 as your stock solution. How many grams of CuSO4(s) should you use?
0.1L * 0.1M = 0.01 moles
Now just convert that to grams.
To determine the number of grams of CuSO4(s) needed to prepare a 0.10 M solution, we need to use the formula:
Molarity (M) = (moles of solute) / (volume of solution in liters)
From the given information, we have:
- Molarity (M) = 0.10 M
- Volume of solution = 100 mL = 0.100 L
Now, let's rearrange the formula to solve for moles of solute:
moles of solute = Molarity (M) * volume of solution (in liters)
Substituting the known values:
moles of solute = 0.10 M * 0.100 L
moles of solute = 0.010 mol
To convert moles to grams, we need to find the molar mass of CuSO4.
Cu has an atomic mass of 63.55 g/mol.
S has an atomic mass of 32.07 g/mol.
O has an atomic mass of 16.00 g/mol.
Adding up the masses for each element:
Cu: 63.55 g/mol
S: 32.07 g/mol
O: (16.00 g/mol * 4) = 64.00 g/mol
The molar mass of CuSO4 is:
63.55 g/mol + 32.07 g/mol + 64.00 g/mol = 159.62 g/mol
Finally, we can calculate the number of grams of CuSO4(s) needed:
grams of CuSO4 = moles of solute * molar mass
grams of CuSO4 = 0.010 mol * 159.62 g/mol
grams of CuSO4 = 1.5962 g
Therefore, you should use approximately 1.60 grams of CuSO4(s) to prepare 100 mL of 0.10 M CuSO4 stock solution.