Question

Name: 1.00g NaCl Volume: 100.34 mL Species (aq) a) The solution labeled 'Solution 1' contains 2.00 g NaCl. How many grams of AgNO3 must be added to the solution to the solution to completely react with NaCl according to the reaction in the background section? Molarity 1.00481e-7 1.00481e-7 0.170529 0.170529 OH Na CI Record the values for the concentration (molarity) for the Nat and Cl' species. 1. Calculate the number of grams of each species and please show work. Name: Solution 1 Volume: 100.68 ml Species (aq) ОН- Take the container containing 1.00 g of AgNO3 and add 100 mL of H20 Name: 1.00g AgNO3 Volume: 100.35 ml Molarity 1.00481e-7 1.00481e-7 0.339907 0.339907 Na Species (s) grams Species (aq) NO, Ag* H Molarity 0.0586647 0.0586647 1.00481e-7 1.00481e-7 OH b) The solution labeled 'Solution 2 contains 3.00 g NaCl. If excess NaCl is added to the solution, how many grams of AgCl solid will be formed? Species (s) grams Obtain the concentrations of the NO3 and Ag+ ions. 2. Calculate the grams of each species and please show work. Name: Solution 2 Volume: 101.05 mL Species (aq) Add the entire amount of the NaCl solution to the flask containing the AgNO3 solution. Name: 1.00g Naci Volume: 200.57 ml OH NO; Agt Molarity 1.00481e-7 1.00481e-7 0.174777 0.174777 Species (aq) Species (s) grams Molarity 1.00481e-7 1.00481e-7 0.0853807 0.0560074 0.0293733 3.27027e-9 Species (s) grams AgCI 0.843704 Record the data for the NO3-, Ag+, Nat, and Cl-ions. 3. Find the grams of AgCl formed and please show work.

Answer 1

of solute dissolved lolasing is defined as no of moles per little of solation. Volume given= 100ml malarity No of moles - U, мне о дəмдім (L) O.IL (16=1000 me) Xb 0.170529 = " ATL 0:170529= 2 No of moles OIL = No of moles = 0.0170529 malasmass of Na=22.9 g mol Males = Weight Molar mass - 0.0170529 =- Weight 22.9 g mol » weight of wat = 0.39051 g for é flatora Settre led due to the tale some molestito moles = 0.170529 =- 0.14 molasmass of cl = 35.4 g mol - moden = 0.0170529 - 0.0170529 - weight 35.48 mol - l weight = 0.6036 g

Ht malapity = 1.00481 x 15 Volume=100 me = O.IL on Molasity = 1.00481x107 volume = olL moles - 1.00481x107 modes > 1.00481 810? oill OIL =) Moles = 1.00481x108 - Moles = 1.00481 x 108 tit makes mass = 108 g mol on = 17.008 g 1.004818 10-8 weight 1.0048x10 8 = weight 1-08 g mol 17.008 g mal » weight = 1.709x107 g af 2) mass of t = 1.085x15 g om