If exactly 14.563g of KMnO4 were dissolved to yield 750mL of solution, what is the normality? I know the equivalent weight is 31.61g. This is how much I got.
(14.563g/.750L)(1 mol KMnO4/158.03g)(5 e-/1 mol) = .615 This answer does not seem right however. If wrong, please explain
The conversion for potassium permanganate, KMnO4, from Normality to Molarity is 1N = 0.2M or 1M = 5N.
Molarity= 14.563/158*.750
now, normality = 5 * molarity above
You are correct.
Okay thank you Bob
To calculate the normality of the KMnO4 solution, you need to know the number of equivalents of KMnO4 present in the solution.
Given:
Mass of KMnO4 = 14.563 g
Volume of solution = 750 mL = 0.750 L
Equivalent weight of KMnO4 = 31.61 g
To calculate the number of equivalents, you can use the formula:
Number of equivalents = (mass of substance / equivalent weight)
Number of equivalents = (14.563 g / 31.61 g)
Number of equivalents = 0.460 equivalents
Now, to calculate the normality of the solution, you need to know the volume of the solution in liters. Since you already have the volume given as 0.750 L, you can use this value.
Normality = (number of equivalents / volume in liters)
Normality = (0.460 equivalents / 0.750 L)
Normality = 0.613 N
So, the normality of the KMnO4 solution is approximately 0.613 N.
In your calculation, it seems like you made an error. Let's correct it:
(14.563g/0.750L)(1 mol KMnO4/158.03g)(5 e-/1 mol) = 0.460
So, you made a calculation error in converting the mass of KMnO4 to the number of equivalents. The correct answer is approximately 0.460 equivalents, which leads to a normality of 0.613 N.