Question

A chemist adds 185.0mL of a 2.6 M iron(II) bromide FeBr2 solution to a reaction flask. Calculate the millimoles of iron(II) bromide the chemist has added to the flask. Round your answer to 2 significant digits.

Answer 1

Answer:

n = 480 millimoles ( 2 significant figures)

Explanation:

Let's bring out the parameters we were given about Iron(II) bromide;

Volume (V) = 185.0ml = 0.185L (Upon conversion to L by dividing by 1000)

Molarity (M) = 2.6

Number of moles (n) = ?

The relationship between these parameters is given a;

The molarity, or concentration of a solution, equals the number of moles in the solution divided by its volume.

The formular is given as;

M = n / V

Making n the subject of interest, we have;

n = M * V

n = 2.6 * 0.185

n = 0.481 moles

Upon conversion to millimoles we have;

n = 481 millimoles

n = 480 millimoles ( 2 significant figures)

Answer 2

Answer:

0.48 moles

Explanation:

The bromide has a molarity of 2.6M.

This simply means that in 1dm^3 or 1000cm^3 of the solution, there are 2.6 moles.

Now, we need to get the number of moles in 185ml of the bromide. It is important to note that the measurement ml is the same as cm^3.

We calculate the number of moles as follows.

If 2.6mol is present in 1000ml

x mol will be present in 185 ml.

To calculate x = (185 * 2.6) ÷ 1000

= 0.481 moles = 0.48 moles to 2 s.f