Answer:
Molarity of the tin(II) chloride solution = 0.3475 M
Explanation:
Iron is present in the wire by a percentage of 94.75% .
Thus,
Percent composition is percentage by the mass of element present in the compound.
Given that the [tex]Fe[/tex] is 94.75 % by mass.
100 g of the wire contains 94.75 g of [tex]Fe[/tex]
0.250 g of the wire contains 0.9475*0.250 g of [tex]Fe[/tex]
Mass of Fe = 0.2369 g
Molar mass of Fe = 55.845 g/mol
The formula for the calculation of moles is shown below:
[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]
Thus,
[tex]Moles= \frac{0.2369\ g}{55.845\ g/mol}[/tex]
Moles of iron = 0.00424 moles
According to the reaction,
[tex]Sn^{2+}_{(aq)}+2Fe^{3+}_{(aq)}\rightarrow 2Fe^{2+}_{(aq)}+Sn^{4+}_{(aq)} + H_2O_{(l)}[/tex]
At equivalence point
Moles of [tex]Sn^{2+}[/tex] = 2*Moles of [tex]Fe^{3+}[/tex]
Considering :-
Moles of [tex]Sn^{2+}=Molarity_{Sn^{2+}}\times Volume_{Sn^{2+}}[/tex]
Volume = 24.4 mL = 0.0244 L ( 1 mL = 0.001 L)
So,
[tex]Molarity_{Sn^{2+}}=\frac{2\times 0.00424}{0.0244}[/tex]
Molarity of the tin(II) chloride solution = 0.3475 M
