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how many grams of iron (III) chloride hexahydrate FeCl3 x 6H2O do you need to make 4 mL of a 1 M FeCl3 solution?

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Final answer:

To make 4 mL of a 1 M FeCl3 solution, 1.0812 grams of iron (III) chloride hexahydrate is required, using the molar mass of 270.3 g/mol for the calculation.

Step-by-step explanation:

To calculate how many grams of iron (III) chloride hexahydrate (FeCl3 x 6H2O) is needed to make a 4 mL of 1 M FeCl3 solution, first, you need to know the molar mass of FeCl3 x 6H2O. Iron (III) chloride hexahydrate has a molar mass of approximately 270.3 g/mol. To make a 1 M solution, you need 1 mole of solute per liter of solution. Since we want to make only 4 mL (0.004 L), we would use:

Mass = Molarity (M) × Volume (L) × Molar Mass (g/mol)

Mass = 1 mol/L × 0.004 L × 270.3 g/mol = 1.0812 g

Therefore, you would need 1.0812 grams of FeCl3 x 6H2O to make 4 mL of a 1 M solution. It's important to note that calculations based on molarity assume that the formation of the solution does not significantly change the total volume, which is a reasonable assumption for such a small amount of substance.

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