Question

How can I obtain molarities of the final solutions with this information that I was told to record in Chem 1 Lab?

AI.

Mass of Cu(CH3COO)2•H2O: 2.495 g

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

Molarity of final solution: ???

AII.

Volume of Solution AI used: 10 mL

Molarity of final solution: ???

BI.

Molarity of stock KMnO4: 0.02011 M

Volume of stock: 20 mL

Molarity of final solution: ???

BII.

Volume of solution BI used: 10mL

Molarity of final solution: ???

CI.

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

Initial buret reading: 15 mL

Final buret reading: 22 mL

Volume of stock solution used: 7 mL

Molarity of final solution: ???

Answer 1

g. 65 A. I. mass of CuCtycoo), 2H, 0 = 2.4959- moles of GCCHg Coo) · H2O = a1991654 g/mol - Modatoily of fivat elutina Volume of solution = lomb.. mol - final solution of o 2.4959 199.654 gemod-x 10x10 = 102(M). = 1:2 mol L CamScanner Scanned with

. B.I Molarity of stock KMnO4 (3) = 0.02011 m. Volume of stock (V.) = 20 ml volume of solution used (V2) = 10 mL .. molarity of final solution zo K x 0.0 2011 m 02 - 2 - some = 0.04022 m, c. I. molarity of stock kz Ch 04 (s.) = 1.25*10-3 m volume of stock solution (%) = 7 ml Initial buret beaching 15 mL Final buset beading 22 mL .. Volume of final solution (^2) = (22-15) mL - 7 mL ... molarity of nervis, 12 final solution A mkx 1.25 x 103 m Amk = 1.25* 10-3 m, Scanned with CamScanner