Question

Revised Question: How can I obtain molarities of the final solutions with this information that I was told to record in Chem 1 Lab? It asks for a molarity of the final solution for EACH part. As in, does (AI) has its own molarity of the solution and (A2) has its own molarity if the final solution or does A1 and A2 have the same molarity of the solution? Same thing for Bl and B2.

Part: AI.

Mass of Cu(CH3COO)2•H2O: 2.495 g

Moles of Cu(CH3COO)2•H2O: 199.654 g/mol

Molarity of final solution: ???

A2.

Volume of Solution AI used: 10 mL

Molarity of final solution: ???

( I got 1.2 M from the data of Al and A2 but is that the molarity of the final solution for both Al and A2?)

*NOTE: My professor did not have us collect the direct data for a volume for Part A1. It was not in our procedures or table. But if it helps once we weighted the mass of Cu(CH3COO)2•H2O = 2.495 g, we then added 25 mL of distilled water to the 50 mL beaker to dissolve it. This is why I got confused when first calculating part A1.

Part:BI.

Molarity of stock KMnO4: 0.02011 M

Volume of stock: 20 mL

Molarity of final solution: ???

B2.

Volume of solution BI used: 10mL

Molarity of final solution: ???

(I got 0.04022 M for the molarity of the final solution from both sets of data but is that the molarity for Bl and B2?)

Part: CI.

Molarity of stock K2CrO4: 1.25 x 10^-3 M

Volume of stock solution used (V1): 7 mL

Volume of stock solution used (V2): 10 mL

Molarity of final solution: ???

( I got the molarity of the final solution = 8.75 x 10^-4 M)

A1 volume=25ml

Answer 1

1-

Given for A1 solution

Volume of solution taken = 25ml

Mass of Cu(CH3COO)2•H2O present = 2.495 g

Again molar mass of Cu(CH3COO)2•H2O = 199.654 g/mol

So mols of Cu(CH3COO)2•H2O present = mass / molar mass

= 2.495 g / 199.654 g/mol

= 0.0125 mols

So molarity of solution = moles of solute / volume of solution in L

= 0.0125 mols / 25ml

= 0.0125 mols / 0.025 L

= 0.5 mols /L

= 0.5 M

Case 1- If we mix solution A1 with another solution A2 with different concentration and volume-

Now if you mix this solution A1 with anothet solution A2 having its own molarity, then the final volume of the addition of tw solution will change

Lets say

Initial molarity of A1 = M1

Initial volume of A1 = V1 mL

Initial molarity of A2 = M2

Initial volume of A2 = V2 mL

Now after addition of the tw, the final volume = V1 + V2 = V3

Then in this final solution, molarity of A1 and A2 will change and-

Molarity of A1 in final solution (lets say M3) = initial molarity * initial volume / final volume

= M1 * V1 / V3

Similarly-

Molarity of A2 in final solution (lets say M4) = initial molarity * initial volume / final volume

= M2 * V2 / V3

Case-2

But if the solution in A2 is nothing but only 10 mL solution taken directly from solution A1

Then in A2, the volume of solution = 10 mL

And molarity of solution = same as A1 = 0.5 M