Elements and Compounds

# What is the molarity of a solution prepared by dissolving 2.41 g of potassium iodide KI in 100 mL of water?

Find moles potassium iodide first.

2.41 grams KI (1 mole KI/166 grams) = 0.01452 moles KI

Molarity = 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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## Related Questions

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The molarity of a solution given by the number of moles divided by the volume it contains. Hence, the molarity of KCl is 4.00/3.00 = 1.33 mol/L.

The moLaRity of this solution ie 2 mol/L or 2M KI (it is a 'two molair potassium iodide' solution)

molarity = no. of moles of solute/liter of solution no. of moles of I2 = mass in grams/molar mass = 4.65/253.81 = 0.01832 mol M = 0.01832 mol/0.235 L = 0.0780 mol/L