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A solution of [tex]$KMnO_4$[/tex] contains 3.16 g of [tex]KMnO_4[/tex] per liter. Find the normality of the solution.

Sagot :

To find the normality of the solution of [tex]\(KMnO_4\)[/tex] containing 3.16 grams of [tex]\(KMnO_4\)[/tex] per liter, we will follow these steps:

1. Calculate the molarity of the solution:
- Determine the molar mass of [tex]\(KMnO_4\)[/tex], which is 158.034 grams per mole.
- Use the formula for molarity:
[tex]\[ \text{Molarity} = \frac{\text{mass of solute (grams)}}{\text{molar mass (grams/mole)} \times \text{volume of solution (liters)}} \][/tex]
- Given that we have 3.16 grams of [tex]\(KMnO_4\)[/tex] in 1 liter of solution, the molarity is:
[tex]\[ \text{Molarity} = \frac{3.16 \text{ grams}}{158.034 \text{ grams/mole}} \][/tex]
- Performing this calculation:
[tex]\[ \text{Molarity} \approx 0.0199957 \text{ moles/liter} \][/tex]

2. Determine the equivalent factor for [tex]\(KMnO_4\)[/tex] in the specific reaction. In acidic medium, [tex]\(KMnO_4\)[/tex] has an equivalence factor of 5 because it gains 5 electrons during the redox reaction.

3. Calculate the normality using the formula:
[tex]\[ \text{Normality} = \text{Molarity} \times \text{Equivalence Factor} \][/tex]
- Substituting the values into the formula:
[tex]\[ \text{Normality} = 0.0199957 \text{ moles/liter} \times 5 \][/tex]
- Performing this calculation:
[tex]\[ \text{Normality} \approx 0.0999785 \text{ equivalents/liter} \][/tex]

So, the normality of the [tex]\(KMnO_4\)[/tex] solution is approximately 0.0999785 equivalents per liter.
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