Question

A chemist prepares a solution of iron(II) bromide FeBr2 by measuring out 0.55kg of iron(II) bromide into a 500.mL volumetric flask and filling the flask to the mark with water.

Calculate the concentration in /molL of the chemist's iron(II) bromide solution.
Round your answer to 2 significant digits

Answer 1

The concentration of the iron(II) bromide solution is 5.10 mol/L, which is calculated by converting the mass of FeBr2 to moles and then dividing by the volume of the solution in liters.

Step-by-step explanation:

To calculate the concentration in moles per liter (mol/L) of the chemist's iron(II) bromide (FeBr2) solution, we need to first convert the mass of FeBr2 into moles. The molecular weight of FeBr2 is approximately 215.65 g/mol. The mass of FeBr2 used is 0.55 kg, which is equivalent to 550 g.

Using the formula:

moles = mass (g) ÷ molar mass (g/mol),

We get:

moles of FeBr2 = 550 g ÷ 215.65 g/mol = 2.550 moles (rounded to 4 significant digits for intermediate calculations).

To find the concentration, we then divide the moles of solute by the volume of the solution in liters:

concentration (mol/L) = moles of solute ÷ volume of solution (L).

The volume of the solution is 0.500 L (since 500 mL is equivalent to 0.500 L). Therefore:

concentration of FeBr2 = 2.550 moles ÷ 0.500 L = 5.10 mol/L (rounded to 2 significant digits).

So, the concentration of the iron(II) bromide solution prepared by the chemist is 5.10 mol/L.

Answer 2

molar mass of ferrous bromide: 215.65
m(FeBr2): 550/215.65 = 2.55moles
this is dissolved in 500mL of water so in 1L of solution there will be 2.55x2 = 5.1mol/L