How many grams of potassium iodide crystals is needed to make 1M.
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To determine how many grams of potassium iodide (KI) crystals are needed to make a 1M (molar) solution, we will need to know the molecular weight of KI and the volume of the solution you want to prepare.
The molecular weight of KI can be found by adding up the atomic weights of its constituent elements. The atomic weight of potassium (K) is approximately 39.1 grams/mole, and the atomic weight of iodine (I) is approximately 126.9 grams/mole. The molecular weight of KI is therefore 39.1 + 126.9 = 166 grams/mole.
A 1M solution contains exactly 1 mole of solute (in this case, KI) dissolved in 1 liter of solvent (usually water).
So, to calculate the number of grams of KI needed for a 1M solution, you can follow these steps:
1. Determine the desired volume of your solution (in liters).
2. Multiply the desired volume by the molarity (1M) to obtain the number of moles you need.
3. Finally, multiply the number of moles by the molecular weight of KI to find the corresponding mass in grams.
For example, if you want to prepare 500 mL (0.5 liters) of a 1M KI solution:
1. Volume = 0.5 liters.
2. Moles = Volume (in liters) × Molarity = 0.5 × 1 = 0.5 moles.
3. Mass (in grams) = Moles × Molecular weight = 0.5 × 166 = 83 grams.
So, in this example, you would need 83 grams of potassium iodide crystals to make a 1M solution in 0.5 liters of solvent.