Question

By pipet, 11.00 mL of a 0.823 MM stock solution of potassium permanganate (KMnO4) was transferred to a 50.00-mL volumetric flask and diluted to the calibration mark. Determine the molarity of the resulting solution. A stock solution of potassium permanganate (KMnO4) was prepared by dissolving 13.0g KMnO4 with DI H2O in a 100.00-mL volumetric flask and diluting to the calibration mark. Determine the molarity of the solution Molarity= O.822 M

Answer 1

Answer:

1) 0.18106 M is the molarity of the resulting solution.

2) 0.823 Molar is the molarity of the solution.

Explanation:

1) Volume of stock solution = $V_1=11.00 mL$

Concentration of stock solution = $M_1=0.823 M$

Volume of stock solution after dilution = $V_2=50.00 mL$

Concentration of stock solution after dilution = $M_2=?$

$M_1V_1=M_2V_2$ ( dilution )

$M_2=\frac{0.823 M\times 11.00 mL}{50 ,00 mL}=0.18106 M$

0.18106 M is the molarity of the resulting solution.

2)

Molarity of the solution is the moles of compound in 1 Liter solutions.

$Molarity=\frac{\text{Mass of compound}}{\text{Molar mas of compound}\times Volume (L)}$

Mass of potassium permanganate = 13.0 g

Molar mass of potassium permangante = 158 g/mol

Volume of the solution = 100.00 mL = 0.100  L ( 1 mL=0.001 L)

$Molarity=\frac{13.0 g}{158 g/mol\times 0.100 L}=0.823 mol/L$

0.823 Molar is the molarity of the solution.