Question

How many milliliters of an aqueous solution of 0.234 M iron(III) bromide is needed to obtain 9.84 grams of the salt? Formulate the question in an open-ended format.

Answer 1

To obtain 9.84 grams of iron(III) bromide, you would need a volume of approximately 42.05 liters of a 0.234 M aqueous solution of the salt.

Step-by-step explanation:

To determine the volume of an aqueous solution of iron(III) bromide needed to obtain a certain mass of the salt, we need to use the equation:

Molarity (M) = moles of solute (iron(III) bromide) / volume of solution (in liters)

First, we need to calculate the moles of iron(III) bromide using the given mass:

Moles of iron(III) bromide = mass of iron(III) bromide / molar mass of iron(III) bromide

Once we have the moles, we can rearrange the equation to solve for the volume of the solution:

Volume of solution (in liters) = moles of solute / Molarity

Plugging in the values, we get:

Volume of solution = 9.84 grams / (0.234 moles/L)

Volume of solution = 42.05 L