What is the molarity of potassium iodide solution prepared by adding 5.810g of KI to sufficient water to make a final solution volume of 250.0mL?
mols KI = grams/molar mass
Then M = mol/L soln.
To find the molarity of a solution, you need to know the moles of solute and the volume of the solution. In this case, the solute is potassium iodide (KI) and the volume is 250.0 mL.
To find the moles of KI, you can use its molar mass. The molar mass of potassium (K) is approximately 39.10 g/mol, and the molar mass of iodine (I) is approximately 126.90 g/mol. Since KI has one potassium and one iodine atom, its molar mass is the sum of the molar masses of the individual atoms:
Molar mass of KI = (molar mass of K) + (molar mass of I)
= (39.10 g/mol) + (126.90 g/mol)
= 166.00 g/mol
Next, you can calculate the moles of KI using its mass and molar mass:
moles of KI = mass of KI / molar mass of KI
= 5.810 g / 166.00 g/mol
≈ 0.0350 mol
Lastly, you can calculate the molarity of the KI solution:
Molarity (M) = moles of solute / volume of solution (in liters)
= 0.0350 mol / 0.2500 L
= 0.140 M
Therefore, the molarity of the potassium iodide solution is approximately 0.140 M.