Question

A chemist dissolved 65.353 grams of CuSO₄ * 5H₂O into enough water to make 150.0 mL of aqueous CuSO₄ solution.  

(a)        Calculate the concentration of this solution, in molarity.

(b)       What is the normality of the Cu²⁺ ions in the above solution?

(c)        He then reacted the above CuSO_{4 solution} with some 0.225 M KI solution. The reaction occurs as follows:      

      2 CuSO₄ (aq) + 4 KI (aq) à 2 CuI (s) + I₂ (aq) + 2 K₂SO₄ (aq)

How many mL of the KI solution will be needed to completely precipitate out all the copper ions of a 15.20 mL of the above CuSO₄ solution?

(d)       How many grams of CuI will he produce in the above titration?

Answer 1

a) x 1000 1000 1500 kn 65-353 1000 15040L of cut2 Icus in soluticy 65. 353 9 of cusou SH₂O SOLUTE Nolarity (M) :- no of moles of solute preseut in IL of solution is called Molarity.CM), Molarity (M) = no ctmoles offolute Volume ct solution in mass of solute Molav massotsolute yin ml 65.353 9 x 248.5 slun 8x mich X 248.5 2. 65353 mork 37425 1:75 M, Concentratich of solution = 1.75m cusoy. 5H₂O » cute+sou IM 1-15m [cut - 1 - 75 M, Normality Molarity & ualency (1.75 M) X2 Normality = 3.5N 2- 1.75m 3.5 N

© of cusou lovo x Ioua 11 0.0266 und zuole of Cui produced, 2 cusoy (aq) +4KI (29) -- 2I(S) + Iz (92) (aa + 2 kz Soy 2 mole of cusou 4 mole of KI is needed. no of moles = molarity x volume in - 1.75 MX 15-20 b = 1.75 mol 15.204 X 0.0266 mol let luc volume of ke soluticy be: VL otmoles ctka no solution = (0.225 M) X VL = (0-225 V) much, 2 mole of cusou needed, o. 02 66 ud og cusay This is equal to 0,225 v ) = 0.2364 236.4mL Volume afkI afkI solution = 2 mole of cusoy 0.0 266 uch of cusoy 4 mole Akilt ~?? no of moles of kI needed 0.0266 X4 2 0.0532 md. 0:0532 0.225 V = 0.0532 0-225 -)?1

no of moles of Cur Cuf precipitated 0.0266 X 27 x 1) 0:0266 un of cu precipitated a Mase no otuolex molarmast 0.0 0266 mphix 90.59 wyn = (0-0266 x 190.5) 9 5. 06739 precipitate tormed Mass of cuI 5-06733