Calculate the molarity of the final solution of Erythrocin B (MW 879.9 g/mol) if a .5028g sample is dissolved in water diluted to 100 mL in a volumetric flask (Solution 1), and then a 5mL aliquot of the stock solution is diluted to 100mL using volumetric equipment. (solution 2)

2.857 × 10⁻⁴ M

Explanation:

First, we will calculate the molarity of the solution 1.

M = mass of solute / molar mass of solute × liters of solution

M = 0.5028 g / 879.9 g/mol × 0.100 L

M = 5.714 × 10⁻³ M

We can find the concentration of the final solution using the dilution rule.

C₁ × V₁ = C₂ × V₂

C₂ = C₁ × V₁ / V₂

C₂ = 5.714 × 10⁻³ M × 5 mL / 100 mL

C₂ = 2.857 × 10⁻⁴ M

explanation: it gains negative electrons, therefore becoming negatively charged. as charge cannot be created, nor destroyed, only transferred, this leaves the wool positive. now that they have opposite charges, the two items attract.