A60.0 ml60.0 ml aliquot of a 1.80 m1.80 m solution is diluted to a total volume of 218 ml.218 ml. a 109 ml109 ml portion of that solution is diluted by adding 107 ml107 ml of water. what is the final concentration? assume the volumes are additive

The final concentration is 0.2497 M.

Explanation:

1. Molarity of the starting solution =

Volume of the starting solution =

After dilution of the solution to 219 mL.

Molarity of solution after dilution =

Volume of the new solution =

(Dilution law)

2. Now,109 mL of 0.495 M solution was diluted by 107 mL

Molarity of the solution =

Volume of the solution =

Molarity of solution after dilution =

Volume of the new solution(107 mL of water is added)  =

(Dilution law)

The final concentration is 0.2497 M.

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%

the electrical  conductivity  of asolution  of an electrolyte is measured by determining the resistance of thesolution  between two flat or cylindrical electrodes separated by a fixed distance.