How many grams of nickel (II) sulfate hexahydrate would be present in one liter of a 0.40 molar solution? (Hint: a 1.0 molar solution would have 1 mole of NiSO4*6H2O per liter of solution)
moles = M x L = ?
grams = moles x molar mass.
625.grams
To find the number of grams of nickel (II) sulfate hexahydrate (NiSO4•6H2O) present in one liter of a 0.40 molar solution, you need to follow these steps:
Step 1: Determine the molar mass of NiSO4•6H2O.
The molar mass of NiSO4 is calculated as follows:
Ni (nickel) = 1 atom x 58.69 g/mol = 58.69 g/mol
S (sulfur) = 1 atom x 32.07 g/mol = 32.07 g/mol
O (oxygen) = 4 atoms x 16.00 g/mol = 64.00 g/mol
H (hydrogen) = 6 atoms x 1.01 g/mol = 6.06 g/mol
Add up the atomic masses: 58.69 g/mol + 32.07 g/mol + 64.00 g/mol + 6.06 g/mol = 160.82 g/mol
Step 2: Calculate the moles of NiSO4•6H2O in one liter of the solution.
Since the solution is 0.40 molar, it means it contains 0.40 moles of NiSO4•6H2O per liter of solution.
Step 3: Convert moles to grams.
Multiply the moles (0.40 moles) by the molar mass of NiSO4•6H2O (160.82 g/mol):
0.40 moles x 160.82 g/mol ≈ 64.33 grams
Therefore, there would be approximately 64.33 grams of nickel (II) sulfate hexahydrate present in one liter of a 0.40 molar solution.