What is the Molarity of Fe(NO3)3? Answer is 360M?

Question

Solution is made by diluting 150 mL of 0.12M Fe(NO3)3 with water to a final volume of 0.5000L.

Answer 1

You've written it down wrong, I think. You've diluted a solution and wound up with one 300 times as concentrated.

As for how to solve it, in my class we used the formula m1v1 = m2v2, or m2 = m1v1/v2, meaning the molarity times the volume is constant.

In this case, m1 is 0.12 M, v1 is 0.150 L and v2 is 0.5000 L. Therefore m2 = 0.12*0.150/0.5000 = 0.0360 M.

Answer 2

First, find the number of moles of Fe(NO3)3 in your original solution.

0.12 mol/L x (1L / 1000mL) x 150 mL = 0.018 moles Fe(NO3)3.

Now obtain your final molarity. 0.018 moles of Fe(NO3)3 in 0.5 L = .018 mol/0.5L = 0.036M.

The major thing wrong with your answer of 360M is that you are not thinking about whether or not it makes sense. If you started with only 0.12M and you are diluting it, that means that your final molarity should be much less than your starting molarity. Always be sure to perform a quick check in this way with any of your answers.

Answer 3

The solution to the question is the dilution formula
M_1V_1=M_2V_2
Therefore,

150*10^-3 L*.12 (mol/L) = .5 L*X (mol/L)
So,
0.018 mol/.5 L = 0.036 M.

Answer 4

150 ml = 0.15 l if the solution is 0.12m
it corresponds to 0.12*0.15 =0.018 mole that you put in 0.5 l

so a liter of that solution
has a molarity of 0.018/0.5 =0.036 M or 36 millimole

And not your value