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We have 495 ml, but we need to convert this to 1000, so we divide by 495 and multiply by 1000. We also do this to 21.1g. 21.1/495*100 is 42.626g. KI's molecular weight is 166g/mol. 42.626/166 is 0.2568 molar.

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How do you prepare a 5 percent potassium iodide solution?

To prepare a 5% potassium iodide solution, weigh 5 grams of potassium iodide and dissolve it in 100 mL of water. Stir until the potassium iodide is completely dissolved to achieve a 5% solution.


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Find moles potassium iodide first.2.41 grams KI (1 mole KI/166 grams) = 0.01452 moles KIMolarity = moles of solute/Liters of solution ( 100 ml = 0.1 Liters )Molarity = 0.01452 moles KI/0.1 Liters= 0.145 M KI solution================


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The solution of potassium iodide (if it is not extremely diluted) is more dense.


How many moles of potassium iodide are needed to make 750 ml of a 1.8 M solution?

To find the number of moles of potassium iodide needed, multiply the volume of the solution (750 ml) by the molarity (1.8 moles/L). First, convert the volume to liters (750 ml = 0.75 L), then multiply 0.75 L by 1.8 moles/L to get 1.35 moles of potassium iodide.


What happens when a solution of potassium iodide is added to a solution of lead nitrate taken in a test tube?

A yellow precipitate of lead iodide is formed due to the reaction between potassium iodide and lead nitrate. This reaction is a double displacement reaction, where the potassium from potassium iodide swaps places with the lead from lead nitrate, forming the insoluble lead iodide.


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