Question

“U” is unknown. The Molarity of CU2+ in volumetric flask (M) is from a graph. each unknown was measured to a final volume of 25.00 mL. Am I moving in the right direction? How would I solve to complete this table?

Solution U-1 U-2 T U-3 U-4 4-5 Absorbance U.OSS 16.136 10.212 10.403 lo.si Molarity of Cuat in volumetric flask (M) 11.0OXIO 250x10 3 4.00810 17.50x 10-3/9.50xloy (determined from graph) Moles or millimoles of Cu?* in volumetric flask 2.50X105 6.25x1os x10-4 1.88810-42.400103 (moles or millimoles) Show calculations 100x10 3 ,02502) -0.00.102.5 2.50x10-5 -0250L ^2 2.56X10-31.0250_) - 0.00 0.0625 625x105 U3 4.00 X10-3 1.0250L) -0.JUL 1.9 U 750 X10-31.0250L) = .000,1,575 TEE 109 Ius 9.50x10 3 1.0250L) = 0007275 2.40X163 Volume of unknown solution usedlosomi 100 ml 150 mL 2.50 m2 300m (mL) Molarity of original unknown Should solution (M) be same Show calculations for all Moles 0.000025 Average molarity of unknown solution: Show calculations for average molarity and avg. dev.

Answer 1

Yes you are moving in the right direction. So just calculate the molarity of original solution using no of moles of cu+2 in volumetric flask by volume of unknown solution by that you can complete the table.

So I provide you the calculations also:

- Molarity nog motel volume of Noume. moles of Certam volumetric folk linknown Santon 2.504 10 1000 o. Soml = 0.05M $ Auster 2.50xlos 6.25 x 10 x2000 for un l.com 1.000 0.0625M .06. 6.2 x 10) x 16oo 1.00x 10 1. Some 1.50 0.066 M 1.oorlory = 0.0752m. "Tu.3 1.88x10 " x1000 2.5ome 2.500 0.08 m 1.888 1044 2.40x10x1000 - 3.oomd 3.00 10 2.468 us