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Sure, I'll guide you through the calculation of the normality of the tin ion in the given solution step by step.
1. Given Data:
- Mass of Tin(IV) chloride, [tex]\(SnCl_4\)[/tex] = 21.1780 grams
- Volume of the solution = 500.0 milliliters (mL)
- Molar Mass of [tex]\(SnCl_4\)[/tex] = 260.52 grams per mole (g/mol)
2. Convert the Volume from mL to Liters (L):
[tex]\[ \text{Volume of solution (L)} = \frac{\text{Volume of solution (mL)}}{1000} \][/tex]
[tex]\[ \text{Volume of solution (L)} = \frac{500.0}{1000} = 0.5 \, \text{L} \][/tex]
3. Calculate the Moles of [tex]\(SnCl_4\)[/tex]:
[tex]\[ \text{Moles of } SnCl_4 = \frac{\text{Mass of } SnCl_4}{\text{Molar Mass of } SnCl_4} \][/tex]
[tex]\[ \text{Moles of } SnCl_4 = \frac{21.1780 \, \text{g}}{260.52 \, \text{g/mol}} \approx 0.08129126362659297 \, \text{mol} \][/tex]
4. Since 1 mole of [tex]\(SnCl_4\)[/tex] produces 1 mole of [tex]\(Sn^{4+}\)[/tex] (because there is one Sn ion in each [tex]\(SnCl_4\)[/tex] unit):
[tex]\[ \text{Moles of } Sn^{4+} = \text{Moles of } SnCl_4 = 0.08129126362659297 \, \text{mol} \][/tex]
5. Determine the Normality:
Normality (N) is defined as the number of equivalents of the solute per liter of solution. For the tin ion [tex]\( (Sn^{4+}) \)[/tex], it undergoes a valence change of 4 in the reaction [tex]\( Sn ^{4+}(aq) + 2e^{-} \rightarrow Sn^{2+}(aq) \)[/tex].
Therefore, the factor of [tex]\( Sn^{4+} \)[/tex] in its half-reaction is 4.
Normality (N) = [tex]\(\frac{\text{Equivalents of solute}}{\text{Volume of solution in liters}}\)[/tex]
The equivalents of [tex]\( Sn^{4+} \)[/tex] will be:
[tex]\[ \text{Equivalents of } Sn^{4+} = \text{Moles of } Sn^{4+} \times 4 \][/tex]
[tex]\[ \text{Equivalents of } Sn^{4+} = 0.08129126362659297 \times 4 = 0.3251650545063719 \][/tex]
Therefore, the Normality (N) is:
[tex]\[ \text{Normality (N)} = \frac{\text{Equivalents of } Sn^{4+}}{\text{Volume of solution in L}} \][/tex]
[tex]\[ \text{Normality (N)} = \frac{0.3251650545063719}{0.5} = 0.6503301090127438 \, \text{N} \][/tex]
So, the normality of the tin ion ([tex]\(Sn^{4+}\)[/tex]) is approximately [tex]\( 0.6503 \, \text{N} \)[/tex].
1. Given Data:
- Mass of Tin(IV) chloride, [tex]\(SnCl_4\)[/tex] = 21.1780 grams
- Volume of the solution = 500.0 milliliters (mL)
- Molar Mass of [tex]\(SnCl_4\)[/tex] = 260.52 grams per mole (g/mol)
2. Convert the Volume from mL to Liters (L):
[tex]\[ \text{Volume of solution (L)} = \frac{\text{Volume of solution (mL)}}{1000} \][/tex]
[tex]\[ \text{Volume of solution (L)} = \frac{500.0}{1000} = 0.5 \, \text{L} \][/tex]
3. Calculate the Moles of [tex]\(SnCl_4\)[/tex]:
[tex]\[ \text{Moles of } SnCl_4 = \frac{\text{Mass of } SnCl_4}{\text{Molar Mass of } SnCl_4} \][/tex]
[tex]\[ \text{Moles of } SnCl_4 = \frac{21.1780 \, \text{g}}{260.52 \, \text{g/mol}} \approx 0.08129126362659297 \, \text{mol} \][/tex]
4. Since 1 mole of [tex]\(SnCl_4\)[/tex] produces 1 mole of [tex]\(Sn^{4+}\)[/tex] (because there is one Sn ion in each [tex]\(SnCl_4\)[/tex] unit):
[tex]\[ \text{Moles of } Sn^{4+} = \text{Moles of } SnCl_4 = 0.08129126362659297 \, \text{mol} \][/tex]
5. Determine the Normality:
Normality (N) is defined as the number of equivalents of the solute per liter of solution. For the tin ion [tex]\( (Sn^{4+}) \)[/tex], it undergoes a valence change of 4 in the reaction [tex]\( Sn ^{4+}(aq) + 2e^{-} \rightarrow Sn^{2+}(aq) \)[/tex].
Therefore, the factor of [tex]\( Sn^{4+} \)[/tex] in its half-reaction is 4.
Normality (N) = [tex]\(\frac{\text{Equivalents of solute}}{\text{Volume of solution in liters}}\)[/tex]
The equivalents of [tex]\( Sn^{4+} \)[/tex] will be:
[tex]\[ \text{Equivalents of } Sn^{4+} = \text{Moles of } Sn^{4+} \times 4 \][/tex]
[tex]\[ \text{Equivalents of } Sn^{4+} = 0.08129126362659297 \times 4 = 0.3251650545063719 \][/tex]
Therefore, the Normality (N) is:
[tex]\[ \text{Normality (N)} = \frac{\text{Equivalents of } Sn^{4+}}{\text{Volume of solution in L}} \][/tex]
[tex]\[ \text{Normality (N)} = \frac{0.3251650545063719}{0.5} = 0.6503301090127438 \, \text{N} \][/tex]
So, the normality of the tin ion ([tex]\(Sn^{4+}\)[/tex]) is approximately [tex]\( 0.6503 \, \text{N} \)[/tex].
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