Question

Prepare 25.00 mL of solution which is 0.60M in isopropanol (0.785 g/mL) and 1.00 M in HCL. What volume (in mL) of isopropanol will you need to add in part II of the experiment?

Also, if a reaction is is second order with respect to isopropanol concentration (and I'm assuming that that would mean the plot of 1/[MnO4 ^2-] vs time is the most linear), what will happen to the rate when you halve the concentration of isopropanol? It would quicken correct or?

Answer 1

of isopropanol, 10 we prepare 25.00 mL 0.60 (M) solution calurate as below : - ull, - moly 0.6 9n 1000 1000 mL no m 1000 9n 25 ml, -0.015 molo w; molewan wt of isob robanol: C3 HD 36 + 8 + 16 60 gm molt E 0.01gx60 Thus: 0.015 moles Density = 0.785 g mit mL ~ 1.15 mL 0.9 0.795 Thus : volume required added needs to be isopropanol Hence; to 25.0 115 ml of mL solvent isopropanol order wirot is and is 2 ; reaction - Given Hence; Rates k [isopropano] is halfea i of concentration Rate' k [isopropanol. Obantu A. (Rate) I from (i) ] Kat u i rate would be slowed down.