Question

What is the final concentration if 35 mL of 0.15 mol/L iron (II) nitrate is mixed with 72mL of a 0.60 mol/L of the same substance? (2 marks)

Answer 1

The final concentration is 0.45mol/L.

Step-by-step explanation:

1st) It is necessary to calculate the moles of iron (II) nitrate contained in 35mL of 0.15 mol/L solution:

`\begin{gathered} 1000mL-0.15moles \\ 35mL-x=(35mL*0.15moles)/(1000mL) \\ x=0.00525moles \end{gathered}`

In this solution there are 0.00525 moles of iron (II) nitrate.

2nd) Now we have to calculate the moles contained in the 72mL of 0.60 mol/L solution:

`\begin{gathered} 1000mL-0.60moles \\ 72mL-x=(72mL*0.60moles)/(1000mL) \\ x=0.0432moles \end{gathered}`

In this solution there are 0.0432 moles of iron (II) nitrate.

3rd) Finally, we have to add both amounts of moles (0.00525 moles + 0.0432moles = 0.04845 moles) considering the total volume (35mL + 72mL = 107mL):

`\begin{gathered} 107mL-0.04845moles \\ 1000mL-x=(1000mL*0.04845moles)/(107mL) \\ x=0.45moles \end{gathered}`

So, the final concentration is 0.45mol/L.