Question

What is the molarity of the solution prepared by diluting 20.0 mL of 0.502 M Mg(NO3)2 solution to each of the following final volumes? a. 40.0 mL Molarity = M b. 125 mL Molarity = c. 1.10 L Molarity = d. 375 mL Molarity = M Submit Answer Try Another Version 10 item attempts remaining

Answer 1

hiver, Initial a For Mg(NO3)2 Initial concentration (Mi) = 0.502M Final Concentration (MF) = ? . Initial volume (V;) = 20 mi = 0.022 (a) Final Volume CVp) = 40 mi = 0.042 we know, "Mi Vi = MFVF 0.502 X 0.02 => ME EVE 0.04 - Mivi = -) Molarity (MF) = 0.251 M - Answer. (b) A VA = 125 mi = 0.125L 0.502 X 0.02 - 0.08 032 => MI = 0.125 VF (Answer) -) ME= 0.08 M (C) VF = HOL Me 0:502 X0.02 - 0.009127 i 1:10 => ME=0.009 M - Answer

(d) 375 ml ie VF = D. 3752 = Me (Molanty) - 7 ME ( Molarity) = Mi Ni - 0.502m x 0.022 VF - 0. 375 ī =) Mf (Molarity) = 0.02 6773 M » Mp = 0.027M 13 decimal places) me round off to Answer