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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)

Respuesta :

Answer:

The final concentration is 0.45mol/L.

Explanation:

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

[tex]\begin{gathered} 1000mL-0.15moles \\ 35mL-x=\frac{35mL*0.15moles}{1000mL} \\ x=0.00525moles \end{gathered}[/tex]

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:

[tex]\begin{gathered} 1000mL-0.60moles \\ 72mL-x=\frac{72mL*0.60moles}{1000mL} \\ x=0.0432moles \end{gathered}[/tex]

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):

[tex]\begin{gathered} 107mL-0.04845moles \\ 1000mL-x=\frac{1000mL*0.04845moles}{107mL} \\ x=0.45moles \end{gathered}[/tex]

So, the final concentration is 0.45mol/L.

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