What mass of lithium iodide is needed to prepare a 0.250 M lithium iodide solution in a 500.0 mL volumetric flask? Express your answer to the correct number of significant figures.
Question
What mass of lithium iodide is needed to prepare a 0.250 M lithium iodide solution in a 500.0 mL volumetric flask? Express your answer to the correct number of significant figures.
Solution
To solve this problem, we need to use the formula for molarity which is M = moles/Liter.
Step 1: Convert the volume from mL to L. 500.0 mL = 0.500 L
Step 2: Calculate the number of moles of lithium iodide (LiI) needed using the molarity formula. Moles of LiI = Molarity x Volume(L) = 0.250 M x 0.500 L = 0.125 moles
Step 3: Convert moles to grams using the molar mass of LiI. The molar mass of LiI is approximately 133.85 g/mol. Mass of LiI = Moles x Molar mass = 0.125 moles x 133.85 g/mol = 16.73 g
So, you would need approximately 16.73 g of lithium iodide to prepare a 0.250 M lithium iodide solution in a 500.0 mL volumetric flask.
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