1. 7.0 mL of a 0.45 M Cu?\" solution was diluted with H20 to a total volume of 1

Question

1. 7.0 mL of a 0.45 M Cu?" solution was diluted with H20 to a total volume of 18.0 ml. What is the molar concentration of Cu2+ molarity Submit Arswer Tries O/2 2. A Beer's law piot for Cu?+ was experimentally obtained. Which graph represents a correct Beer's law plot? Conc Conc Conc Conc Suomit Arswer Tries 0/2 rintercht-0.001). A at'solution of unknown concentration had an 3. A Beer's law plot for Cu was experimentally obtained. The slope of the Beer's law plot was 435 L/mol absorbance of 0.95. What is the molar concentration of Cu2+ in the unknown solution? Cu2 molarity Submit Answer Tries 0/2 Use these 2 reactions to answer Question 4: Cu?+ solution of unknown concentration was placed in a 250 mL Erlenmeyer fiask. An excess of KI solution was added. Indicator was added 4. For rxn 1, 10.0 mL diluted with H20 to a total volume of 75 mL. For man 2, the solution from ran 1 was titrated with 0.10 M Na 5,03. The equivalence point of the titration was reached when 11.30 ml of Na25)0 hod been added. What is the molar concentration of Cu2+ in the unknown solution? of

Explanation / Answer

moles of Cu+2 in 7 ml of 0.45M= Molarity* volumer in L, 1000ml =1L

Molarity of Cu+2= 0.45*7/1000

this has been diluted to 18ml. During dilution, moles of Cu+2 remain the same. Hence

0.45*7/1000 =x* 18/1000

x= molarity after dilution = 0.45*7/18=0.175 M

2, Beers law equation is A(absorbance)= e(molar extinction coefficient)*b*(path length)* C( concentration)

S o the plot of A vs C has to be straight line ( So B is correct)

3. Beers law equation is A= m* concentration + intercept

m= slope

Hence 0.95= 435* concentration+0.001

concentration = (0.95-0.001)/435=0.002182 mol/L

4. let x= concentration of Cu+2, mole of Cu+2 in 10ml= Molarity*Volune in Liters= x*10/1000 moles

the reaction -1 is 2Cu+2 + 4KI------>2CuI+I2+4K+

2 mole of Cu+2 forms 1 mole of I2 with excess of KI

x*10/1000 mole of Cu+2 forms x*10/2000 moles of I2

as per the reaction-2

I2+2Na2SO3------->2NaI+Na2S4O6

moles of Na2SO3 used = 11.3*0.1/1000 = moles of I2 formed from reaction -1 =x*10/2000

x= 11.3*0.1*2/10= 0.126M

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