How much of solid potassium chromate whould you use to prepare 2.00 L of 0.400 M potassium chromate?
M x L = mols
g/molar mass = mols OR
M x L = g/molar mass
solve for g.
.400 mol/L x 2.00 = g/194
g = 155.2
Thanks!!
To prepare 2.00 L of a 0.400 M potassium chromate solution, we need to calculate the amount of solid potassium chromate needed.
First, we can use the formula Molarity (M) x Volume (L) = moles (mol) to find the number of moles required.
0.400 mol/L x 2.00 L = 0.800 moles.
Next, we can use the formula grams (g) = moles (mol) x molar mass (g/mol) to calculate the amount of potassium chromate in grams.
The molar mass of potassium chromate (K2CrO4) is 194 g/mol.
0.800 moles x 194 g/mol = 155.2 grams.
Therefore, you would need to use 155.2 grams of solid potassium chromate to prepare 2.00 L of a 0.400 M potassium chromate solution.