If 5.35 g of KMnO4 are dissolved in a liter of solution, what is the normality of the solution?
To find the normality of a solution, we need to know the number of equivalents of the solute present in one liter of the solution.
First, we need to determine the number of moles of KMnO4 present in the solution. To do this, we divide the given mass of KMnO4 (5.35 g) by its molar mass. The molar mass of KMnO4 is calculated as follows:
(1 x atomic mass of K) + (1 x atomic mass of Mn) + (4 x atomic mass of O)
= (1 x 39.10 g/mol) + (1 x 54.94 g/mol) + (4 x 16.00 g/mol)
= 39.10 g/mol + 54.94 g/mol + 64.00 g/mol
= 158.04 g/mol
Now let's calculate the number of moles:
Number of moles = mass / molar mass
Number of moles = 5.35 g / 158.04 g/mol ≈ 0.0338 mol
Next, we need to determine the number of equivalents of KMnO4 present. The equivalent weight is calculated by dividing the molar mass by the number of equivalents. In the case of KMnO4, each formula unit contributes one equivalent of the permanganate ion (MnO4-).
Equivalent weight = molar mass / number of equivalents
Equivalent weight = 158.04 g/mol / 1 equivalent/mol
Equivalent weight = 158.04 g/mol
Now that we know the number of equivalents of KMnO4 in the solution is the same as the number of moles, we can calculate the normality. Normality is defined as the number of equivalents per liter of solution:
Normality = number of equivalents / volume of solution (in liters)
Since the volume of solution is given as 1 liter, the normality is equal to the number of equivalents of KMnO4 present in 1 liter of the solution, which is approximately 0.0338 N.