A chemistry teacher needs to prepare 750 ml of a 2.5M solution of Potassium iodide for a lab. How many grams of potassium iodide will be needed to make the solution?
.75 * 2.5 * 166
To find the number of grams of potassium iodide needed to make a 2.5M solution of potassium iodide, we can use the formula:
moles = concentration (M) × volume (L)
First, we need to convert the given volume from milliliters (ml) to liters (L):
750 ml ÷ 1000 = 0.75 L
Now, we can use the formula to calculate the number of moles of potassium iodide required:
moles = 2.5 M × 0.75 L
= 1.875 moles
To convert moles to grams, we need to know the molar mass of potassium iodide. The molar mass of potassium iodide (KI) is calculated by adding the atomic masses of potassium (K) and iodine (I):
atomic mass of K = 39.10 g/mol
atomic mass of I = 126.90 g/mol
molar mass of KI = 39.10 + 126.90 = 166.00 g/mol
Finally, we multiply the number of moles by the molar mass to find the mass in grams:
mass (g) = moles × molar mass
= 1.875 moles × 166.00 g/mol
≈ 310.87 g
Therefore, approximately 310.87 grams of potassium iodide will be needed to make the 750 ml (0.75 L) solution with a concentration of 2.5M.