You wish to make a 0.200 M solution of NiSO4 (aq). How many grams of NiSO4 * 6H2O should you put in a 0.500 L volumetric flask?
The answer is 26.3 g NiSO4 * 6H20 but I am not sure how to get there.
Find moles first:
That is M x L = 0.20 x 0.500L = ?
Then convert from moles to grams with moles/molar mass. You want to use the molar mass of NiSO4.6H2O:
0.10 mol x 262.84 g/mol = 26.28 g of NiSO4.6H2O
To calculate how many grams of NiSO4 * 6H2O to put in the volumetric flask, you need to use the molarity (M) and volume (V) relationship. Here's how you can calculate it step by step:
1. Determine the molar mass of NiSO4 * 6H2O:
- NiSO4: Ni (58.69 g/mol) + S (32.07 g/mol) + 4O (16.00 g/mol) = 154.75 g/mol
- H2O: 2H (1.01 g/mol) + 16O (16.00 g/mol) = 18.02 g/mol
- Multiply the molar mass of H2O by 6 since there are 6 water molecules per NiSO4 * 6H2O: 18.02 g/mol * 6 = 108.12 g/mol
- Add the molar masses of NiSO4 and 6H2O: 154.75 g/mol + 108.12 g/mol = 262.87 g/mol
2. Determine the moles of NiSO4 needed using the molarity and volume relationship:
- Molarity (M) is given as 0.200 M.
- Volume (V) is given as 0.500 L.
- Moles (n) = Molarity (M) * Volume (V): 0.200 M * 0.500 L = 0.100 mol
3. Calculate the grams of NiSO4 * 6H2O needed based on the determined moles:
- Grams (m) = Moles (n) * Molar mass (M): 0.100 mol * 262.87 g/mol = 26.3 g
Therefore, you should put 26.3 grams of NiSO4 * 6H2O in the 0.500 L volumetric flask to make a 0.200 M solution.