) g /mol = 3.86!10"3 moles

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1 Chem01, Winter 006 Midterm N1 01/6/06 Name Answer key SID 1. A solution is prepared by dissolving 5.8 grams on magnesium chloride (MgCl ) in water to produce 50.0 ml of solution. Molecular weight of the MgCl is 95.3 g/mol. Molecular weights of Mg and Cl are 4.3 g/mol and 35.5 g/mol, respectively. a. Calculate the molarity of the chloride ion in the solution. (3points) n Cl- = n MgCl = (m MgCl/ MW MgCl) = moles M Cl- = n Cl- / Volume = moles / 0.5 L =.17 M b. What is the concentration of the Cl - in ppm? (3points) Cl - ppm = mass Cl - (mg) / volume = n Cl- MW Cl- / volume = x 35.5 x 1000 / 0.5L = ppm = 7.68 x 10 4 ppm c. Calculate the pcl - value for this solution. (3points) pcl - = -log [.17] = A bottle of a concentrated aqueous sulfuric acid is, labeled 98.0 wt % H SO 4 (Molecular weight is g/mol) has a concentration of 18.0 M. a. How many milliliters of reagent should be diluted to L to give 1.00 M H SO 4? (5 points) V con = V dil x (M dil / M con ) = 1000 ml x ( 1.00 M / 18.0 M) = 55.6 M b. Calculate the density of 98.0 wt % H SO 4 (5 points) Mass of the 1 liter of H SO 4 : (18 moles) (98.09) = 1.77 x 10 3 gr. Mass of the 1 ml of H SO 4 : 1.77 g d = mass / weight % = 1.77 g / (0.98 g H SO 4 /g solution) = 1.8 g/ml 3. How many milliliters of 3.00 M sulfuric acid are required to react with 4.35 g of solid containing 3. g wt % Ba(NO 3 ) if the reaction is: Ba + + SO 4 - BaSO 4? (5 points) Molecular weights of BaSO 4 is 33.0 g/mole and Ba(NO 3 ) is 61.3 g/mol. Mass Ba(NO 3 ) is 0.3 x 4.35 = 1.01 g moles Ba + = mass Ba(NO ) g = MW Ba(NO 3 ) 61.34g /mol = 3.86!10"3 moles moles H SO 4 = moles Ba + = 3.86!10 "3 moles

2 volume H SO 4 = moles H SO 4 M H SO 4 = 3.86!10"3 moles 3 moles/l =1.9!10 "3 L =1.9 ml 4. A sample is certified to contain 94.6 ppm of a contaminant. Your analysis gives values of 98.6, 98.4, 97., 94.6 and 96. ppm. Do you results differ from the expected result at following confidence levels: i) 95%, ii) 99% and iii) 99.9%. (9points) x = 97.0 s = 1.65 = "(! x) n!1 t calc = µ! x s n = 94.6! = 3.5 t table 95% =.776 < 3.5 Significant difference t table 99% = > 3.5 t table 99.9% = > 3.5 No significant difference No, significant difference 5. Using the appropriate statistical test, decide whether the value 16 should be rejected from the set of result: 19, 16, 0, 195 and 04? (3 points) Gap = 1 Range = 4 Q calc = Gap Range = 1 4 = 0.5 < Q table = 0.64 Value to be retained. 6. The following data was collected when performing a spectrophotometric analysis for cobalt. x y Analysis No mg Co / liter Absorbance a. Using the least squares method of linear regression, generate the equation to define the line for the absorbance vs. concentration. (10 points).! = = 31.16

3 ! y i = = 0.588! x i = (5.3) + (10.5) + (15.41) = ! y i = (5.3" 0.095) + (10.5" 0.198) + (15.41" 0.95) = 7.13 n = 3 D = x i n = =155.5 m = x y i i y i n D = = b = x i y i D = y i =! Thus, linear regression line is: y = x! b. Based on the equation you have generated, calculate the concentration of the Co in the sample if the absorbance is: i) ( points) x = 0.155! (!0.0089) ii) 0.65 ( points) x = 0.65! (!0.0089) = 8.3 mg/l =13.90 mg/l 7. Chloroform is an internal standard in the determination of the pesticide DDT in a polarographic analysis. A mixture containing mm chloroform and mm DDT gave signals of 15.3 A for chloroform and 10.1 A for DDT. An unknown solution (10.0 ml) containing DDT was placed in a ml volumetric flask and 10. L of chloroform (FW g/mol, density = g/ml) were added. After diluting to the mark with solvent, polarographic signals of 9.4 and 8.7 A were observed for the chloroform and DDT, respectively. Find the concentration of DDT in unknown. (10 points) Chloroform is S, and DDT is X: A x X = F A s S! 10.1µA 15.3 µa = F mm mm! F = 0.41 Concentration of the chloroform in unknown:

4 (10.10 "6 L) (1484 g/l) / g/mol = M =1.6 mm L For the unknown mixture: A x X = F A s 8.7 µa! S X 9.4 µa = mm! X = mm DDT in unknown: 100 ml mm! = 9.09 mm 10 ml 8. A beaker contains 50.0 ml of molar silver ion (Ag + ). To this beaker is added 50.0 ml of molar bromide ion (Br - ). What is the concentration of Ag + in the final solution? for the AgBr is ? (5 points) AgBr! Ag + + Br " Final concentration of the Ag + and Br - : [Ag + ] = M [Br! ] = M = M = M Br - ion is in excess: = M. [Ag + ] = x and [Br - ] = (x+0.075) = (x)(x ) = 5.0!10 "13 Assuming x << 0.075, we have = (x) (0.075) = 5.0!10 "13 x = [Ag + ] = Iron in the + oxidation state reacts with potassium dichromate to produce Fe 3+ and Cr 3+ according to the equation: 6 Fe + + Cr O H + 6 Fe 3+ + Cr H O How many milliliters of molar K Cr O 7 are required to titrate 00.0 ml of molar Fe + solution? (5 points) n Fe + = 6! n Cr O 7 " # M Fe +V Fe + = 6M Cr O 7 "V Cr O 7 " Therefore,

5 V Cr O 7 = M Fe + "V Fe + 6 " M Cr O 7 = M " 00.0 ml 6" M = 50.6 ml 10. A mixture having a volume of 10.0 ml and containing M Ag + and M Hg + was titrated with M KCN to precipitate Hg (CN) ( = ) and AgCN ( = ). Calculate the concentration of the CN - at each of the following volumes of added KCN: Hg + + CN! " Hg (CN) Ag + + CN! " AgCN Hg + will precipitate first and the equivalence point is at 0.00 ml. And the second equivalence point is at 30 ml. At 5 and 15 ml there is an excess of unreacted Hg +. a ml (5points) " [Hg + ] = 0! 5 % $ ' # 0 & ( ) " 10 % $ ' = M # & [CN! ] = [Hg + ] = 5 "10! =1.0"10!19 M b ml (5points) " [Hg + 0!15% ] = $ ' # 0 & ( ) " 10 % $ ' = M # & [CN! ] = [Hg + ] = 5"10! =.3"10!19 M c ml (5points) At ml, there are 5 ml excess of the [CN - ]: " 5.00 % [CN! ] = $ ' # & ( ) = M 11. Calculate the concentration of Ag + in saturated solutions of Ag CO 3 ( = ) in: + 3 K = [ Ag ] " x " " sp Ag + [ CO3 ]" = x " + x" = CO3 Ag CO3 Ag + CO 3

6 x x x = [Ag + ] = 3! Ag +! " CO3 Corresponding activity coefficients are taken from table (see supplemental information). (a) M KNO 3 (5points) ( ) = µ = 1! c i z = 1 i (0.001"1 ) + (0.001"1 ) [Ag + ] = 3 =! Ag +! " CO3 8.1# = 0.16 mm (0.964) (0.867) (b) 0.01 M KNO 3 (5points) ( ) = 0.01 µ = 1! c i z = 1 i (0.01"1 ) + (0.01"1 ) [Ag + ] = 3 =! Ag +! " CO3 8.1# = 0.47 mm (0.898) (0.665) (c) 0.1 M KNO 3 (5points) ( ) = 0.1 µ = 1! c i z = 1 i (0.1"1 ) + (0.1"1 ) [Ag + ] = 3 =! Ag +! " CO3 8.1# = mm (0.75) (0.37)

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