A stock solution of potassium permanganate (KMnO4) was prepared by dissolving 11.3 g KMnO4 with DI H2O in a 100.00-mL volumetric flask and diluting to the calibration mark. Determine the molarity of the solution.

[KMnO₄] = 0.71 M

Explanation:

Solute: Potassium permanganate, molar mass 158.04 g/mol

Solvent: Water, volume of 100 mL

We convert the mass of solute to moles:

11.3 g . 1mol / 158.04g = 0.071 moles

We asume the solvent's volume as the solution's volume

We convert the volume from mL to L → 100 mL . 1L / 1000 mL = 0.1L

Molarity (mol/L) → moles of solute in 1L of solution

Molarity → 0.071 moles / 0.1 L = 0.71M

percent abundance for cu-63 can be shown as:

63 x + 65 (1-x) = 62.9296

65 - 62.9296 = 2 x

x = 1.035

thus, percent abundance for cu-63 is 1.035

percent abundance for cu-65 can be shown as:

63 x + 65 (1-x) = 64.9278

65 - 64.9278 = 2 x

x = 0.0361

thus, percent abundance for cu-63 is 0.0361

the mass spectrum of cu-63 showing the intense peak as, this isotope is found in nature more abundantly around 92.38 % in comparison to cu-65 which is found 62.58 % naturally