X Xavy New member Dec 12, 2009 #1 What is the molarity of a solution composed of 2.08 g of potassium iodide (KI) dissolved in enough water to make 0.393 L of solution? Answer in units of M.
What is the molarity of a solution composed of 2.08 g of potassium iodide (KI) dissolved in enough water to make 0.393 L of solution? Answer in units of M.
N Norrie New member Dec 12, 2009 #2 Mol.mass of KI = 39 + 127 = 166g/mol. 0.393L x 1,000 = 393mL. ((2.08g / 166g/mol) / 393mL)) x 1,000mL/L = 0.0319 M.
Mol.mass of KI = 39 + 127 = 166g/mol. 0.393L x 1,000 = 393mL. ((2.08g / 166g/mol) / 393mL)) x 1,000mL/L = 0.0319 M.
W William G New member Dec 12, 2009 #3 Step 1: Find Molar Mass of Kl. Kl = 39.10 + 126.9 = 166g Step 2: Find the moles of KI 2.08g KI (1 mol KI / 166g KI) = .0125 mol KI Step 3: Using Molarity formula find molarity. Molarity = mole of Solute / L of Solution = .0125 mol KI / 0.393 L = .0318 M Bill
Step 1: Find Molar Mass of Kl. Kl = 39.10 + 126.9 = 166g Step 2: Find the moles of KI 2.08g KI (1 mol KI / 166g KI) = .0125 mol KI Step 3: Using Molarity formula find molarity. Molarity = mole of Solute / L of Solution = .0125 mol KI / 0.393 L = .0318 M Bill
