Respuesta :
Answer : The number of Fe(II) ions present are [tex]5.94\times 10^{22}[/tex]
Explanation : Given,
Mass of [tex]FeSO_4[/tex] = 15.0 g
Molar mass of [tex]FeSO_4[/tex] = 152 g/mole
First we have to calculate the moles of [tex]FeSO_4[/tex]
[tex]\text{Moles of }FeSO_4=\frac{\text{Mass of }FeSO_4}{\text{Molar mass of }FeSO_4}=\frac{15.0g}{152g/mole}=0.0987mole[/tex]
Now we have to calculate the number of Fe(II) ions.
In [tex]FeSO_4[/tex], there 1 atom of iron ion and 1 atom of sulfate ion.
As we know that,
1 mole of substance always contains [tex]6.022\times 10^{23}[/tex] number of atoms or ions.
As, 1 mole of [tex]FeSO_4[/tex] contains [tex]6.022\times 10^{23}[/tex] number of Fe(II) ions.
So, 0.0987 mole of [tex]FeSO_4[/tex] contains [tex]0.0987\times 6.022\times 10^{23}=5.94\times 10^{22}[/tex] number of Fe(II) ions.
Therefore, the number of Fe(II) ions present are [tex]5.94\times 10^{22}[/tex]
