Question

33) If 100. mL of water is added to 25 mL of 5.0 M NaCl solution, what is the final concentration? 34) Calculate the volume of 0.123 M AgNO3(aq) needed to form 0.657 g of Ag2SO4(s) according to the chemical equation: 2AgNO3(aq) + H2SO4(aq) -----> Ag2SO4(s) + 2H20(1). 35) Calculate the concentration of a KCl solution prepared by adding 200. mL of water to 100 mL of 12% KCl solution.

Answer 1

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33)

(33) SO Molarite = No. of moles volo of solution (l) Initially, No. of moles of Nacd = Molarity of Nace & Volume (1) = 5(M) X 0.025 (e) = 0.125 mole On adding additional water volume of soll changes, but moles of Nucl remain same, any.e. 0.125 mole Volume of soln after adding 100 me of water = 0.100 + 0.025(1) = 0.125 e so, final concentration of or al Molarity = solu mole of Wall mob at Noul final role of sorres 0.125 - - 1 0.125

34)

-39 -3 = 2IIXD mole The Chemical Equi- 2 AgNO3 Cap.) + H₂SO (98) Ag, so (s) + 2Hore). Crineu wto of Ag2 soa = 0.657 g. Moleculon wt of Ag Soo = 311g. Mole of Ginen. Wt. 0.657 Ag2 so 4 = molecular wt. from balanced &gh: - for 1 mole of Ag, soe > 2 moles of Ag Noz is required. for 2011 x 103 mole of Aq sog (2x 2- 11 x 10²3) of Ag No, is required. = 4.224103 mole of Ag was molarity of Ag No = more of AgNO3 = 0.123 m (given) Vol. of AgNO3 (2) - Volo of Agno (ind) = 4.22x10 3 more 0.123 m = 34.3 ml mole

35)

35 Initial volume of solution ( = 100 me final vole of solution (V2) = 100 ml + 200 me = 300 ml Initial concentration of kee = 12 os. = M1 final concentration of kce = M₂ = ?? Mal2 = MOV (mole of kel semains same) . - Me= Movi Vo - (12.). ) X 100 ml 300 ml final Concentration M₂ = 14%. of kee solution. so, PLEASE UPROTE