Final answer:
The volume of a 1.47M iron(II) bromide solution containing 25.0g of iron(II) bromide is found to be 78.8 mL after converting grams to moles and using the molarity equation. The answer is not among the provided options a) through d), suggesting a mistake in the question or options.
Step-by-step explanation:
To calculate the volume in milliliters of a 1.47M iron(II) bromide solution that contains 25.0g of iron(II) bromide, we need to use the molarity equation M = n/V, where M is molarity, n is the number of moles, and V is the volume in liters. We first convert the grams of iron(II) bromide to moles using the molecular weight of iron(II) bromide (FeBr2), which is approximately 215.65 g/mol.
Number of moles n = mass (g) / molar mass (g/mol) = 25.0 g / 215.65 g/mol = 0.1159 moles
Now, we can calculate the volume V, using the molarity equation rearranged to V = n/M.
Volume V = n / M = 0.1159 moles / 1.47 mol/L = 0.0788 L
Converting liters to milliliters (1 L = 1000 mL), we get:
Volume V = 0.0788 L × 1000 mL/L = 78.8 mL
The correct answer is not provided in the options a) through d), so there appears to be a mistake in the question or the provided options.