calculate the mass of Cuso4.5H2O required to prepare 500 cm3 of a 0.400 m/dm-3 solution
Assuming you meant 0.400g/dm^3, that translates to 0.0004 g/cm^3
So, 500cm^3 of that means 0.2 g
Or, in one step, that's
500 cm^3 * (1dm/10cm)^3 * 0.4 g/dm^3
I think 0.400 m/dm3 meant 0.400 mols/dm3 = 0.400 M.
To calculate the mass of CuSO4.5H2O required to prepare a 0.400 M/dm^3 solution, we need to use the molarity formula:
Molarity (M) = moles of solute / volume of solution (dm^3)
Given:
Desired molarity (M) = 0.400 M
Volume of solution (V) = 500 cm^3 = 500/1000 dm^3 = 0.500 dm^3
We need to convert the molarity to moles. The molar mass of CuSO4.5H2O is required for this calculation. Let's break down the molar mass:
- CuSO4:
- Atomic mass of Cu = 63.546 g/mol
- Molecular mass of SO4 = 32.06 g/mol
- Total mass of CuSO4 = (63.546 + 32.06 * 4) g/mol = 159.608 g/mol
- H2O:
- Atomic mass of H = 1.008 g/mol
- Atomic mass of O = 16.00 g/mol
- Total mass of H2O = (1.008 * 2 + 16.00) g/mol = 18.016 g/mol
Now, let's calculate the mass of CuSO4.5H2O:
- Moles of CuSO4.5H2O = Molarity * Volume
- Moles of CuSO4.5H2O = 0.400 M/dm^3 * 0.500 dm^3
- Moles of CuSO4.5H2O = 0.200 moles
The molar mass of CuSO4.5H2O is the sum of the individual masses:
Molar mass of CuSO4.5H2O = (159.608 g/mol + 18.016 g/mol) g/mol
Molar mass of CuSO4.5H2O ≈ 177.624 g/mol
Now, to find the mass of CuSO4.5H2O needed, we can use the formula:
Mass = Moles of solute * Molar mass
Mass = 0.200 moles * 177.624 g/mol
Mass ≈ 35.525 g
Therefore, approximately 35.525 grams of CuSO4.5H2O are required to prepare 500 cm3 of a 0.400 M/dm^3 solution.
To calculate the mass of CuSO4·5H2O required to prepare a 0.400 m/dm³ solution, we'll need to follow these steps:
Step 1: Determine the molar mass of CuSO4·5H2O.
CuSO4·5H2O consists of a copper ion (Cu2+), a sulfate ion (SO4²⁻), and five water molecules (H2O).
The molar mass of copper (Cu) is 63.55 g/mol.
The molar mass of sulfur (S) is 32.06 g/mol.
The molar mass of oxygen (O) is 16.00 g/mol.
The molar mass of hydrogen (H) is 1.01 g/mol.
The molar mass of water (H2O) is (2 x 1.01 g/mol) + 16.00 g/mol = 18.02 g/mol.
Calculating the molar mass of CuSO4·5H2O:
(1 x Cu molar mass) + (1 x S molar mass) + (4 x O molar mass) + (10 x H molar mass) = (1 x 63.55 g/mol) + (1 x 32.06 g/mol) + (4 x 16.00 g/mol) + (10 x 1.01 g/mol) = 249.68 g/mol.
Step 2: Calculate the number of moles of CuSO4·5H2O required.
Molarity (M) is defined as moles of solute per liter of solution (mol/dm³). In this case, we have a 0.400 m/dm³ solution, meaning there are 0.400 moles of CuSO4·5H2O in 1 liter of solution (or 1000 cm³).
Therefore, the number of moles of CuSO4·5H2O required for 500 cm³ (0.500 dm³) of the 0.400 m/dm³ solution is:
0.400 mol/dm³ × 0.500 dm³ = 0.200 moles of CuSO4·5H2O.
Step 3: Calculate the mass of CuSO4·5H2O required.
Finally, we can calculate the mass of CuSO4·5H2O needed using the calculated number of moles and the molar mass of CuSO4·5H2O:
Mass = Number of moles × Molar mass
Mass = 0.200 moles × 249.68 g/mol
Mass = 49.94 grams
Therefore, you would need 49.94 grams of CuSO4·5H2O to prepare 500 cm³ of a 0.400 m/dm³ solution.