Posted by donald on .
Calculate the grams of K2CO3 required to prepare 546.0 mL of a 0.789 g/L K+ solution.
moles K^+ needed = M x L = ?
Convert moles K ions to mol K2CO3.
moles K ion x (1 mol K2CO3/2 mol K ion) = ?
Then moles K2CO3 = grams/molar mass.