Download Chemistry Midterm Exam Solutions, Winter 2006 and more Lecture notes Chemistry in PDF only on Docsity!
Chem201, Winter 2006 Name Answer key ______________ Midterm N 01/26/ SID___________________________
- A solution is prepared by dissolving 25.8 grams on magnesium chloride (MgCl 2 ) in water to produce 250.0 mL of solution. Molecular weight of the MgCl 2 is 95.3 g/mol. Molecular weights of Mg and Cl are 24.3 g/mol and 35.5 g/mol, respectively.
a. Calculate the molarity of the chloride ion in the solution. (3points)
n (^) Cl- = 2 n (^) MgCl2 = 2 (m (^) MgCl2 / MW (^) MgCl2 ) = 0.541 moles
M (^) Cl- = n (^) Cl- / volume = 0.541 moles / 0.25 L = 2.17 M
b. What is the concentration of the Cl -^ in ppm? (3points)
Cl -^ ppm = mass Cl -^ (mg) / volume = n (^) Cl- MW (^) Cl- / volume = 0.541 x 35.5 x 1000 / 0.25L =
76840 ppm = 7.68 x 10 4 ppm
c. Calculate the pCl -^ value for this solution. (3points)
pCl -^ = -log [2.17] = -0.
- A bottle of a concentrated aqueous sulfuric acid is labeled 98.0 wt % H 2 SO 4 (Molecular weight is 98.09 g/mol) has a concentration of 18.0 M.
a. How many milliliters of reagent should be diluted to 1.000 L to give 1.00 M H 2 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 2 SO 4 (5 points)
Mass of the 1 liter of H 2 SO 4 : (18 moles) (98.09) = 1.77 x 10 3 gr. Mass of the 1 mL of H 2 SO 4 : 1.77 g
d = mass / weight % = 1.77 g / (0.98 g H 2 SO 4 /g solution) = 1.8 g/mL
- How many milliliters of 3.00 M sulfuric acid are required to react with 4.35 g of solid containing 23.2 g wt % Ba(NO 3 ) 2 if the reaction is:
Ba 2+^ + SO 4 2-^ BaSO 4? (5 points)
Molecular weights of BaSO 4 is 233.0 g/mole and Ba(NO 3 ) 2 is 261.3 g/mol.
Mass Ba(NO 3 ) 2 is 0.232 x 4.35 = 1.01 g
- A sample is certified to contain 94.6 ppm of a contaminant. Your analysis gives values of 98.6, 98.4, 97.2, 94.6 and 96.2 ppm. Do you results differ from the expected result at following confidence levels: i) 95%, ii) 99% and iii) 99.9%. (9points)
s = 1.65 =
Significant difference
No significant difference
No, significant difference
- Using the appropriate statistical test, decide whether the value 216 should be rejected from the set of result: 192, 216, 202, 195 and 204? (3 points)
Gap = 12 Range = 24
Value to be retained.
chloroform and DDT, respectively. Find the concentration of DDT in unknown. ( points) Chloroform is S , and DDT is X :
Concentration of the chloroform in unknown:
For the unknown mixture:
DDT in unknown:
- A beaker contains 250.0 mL of 0.150 molar silver ion (Ag +^ ). To this beaker is added 250.0 mL of 0.300 molar bromide ion (Br -^ ). What is the concentration of Ag +^ in the final solution? K (^) sp for the AgBr is 5.0 10 -13^? (5 points)
Final concentration of the Ag +^ and Br -^ :
Br -^ ion is in excess: 0.150 – 0.075 = 0.075 M. [Ag +^ ] = x and [Br -^ ] = (x+0.075)
Assuming x << 0.075, we have
x = [Ag +^ ] = 6.67 10 -^.
- Iron in the +2 oxidation state reacts with potassium dichromate to produce Fe 3+^ and Cr 3+ according to the equation:
6 Fe 2+^ + Cr 2 O 7 2-^ + 14 H +^ 6 Fe 3+^ + 2 Cr 3+^ + 7 H 2 O How many milliliters of 0.1658 molar K 2 Cr 2 O 7 are required to titrate 200.0 mL of 0. molar Fe 2+^ solution? (5 points)
Therefore,
- A mixture having a volume of 10.0 mL and containing 0.100 M Ag +^ and 0.100 M Hg 2 2+ was titrated with 0.100 M KCN to precipitate Hg 2 (CN) 2 (K (^) sp = 5 10 -40^ ) and AgCN (K (^) sp = 2.2 10 -16^ ). Calculate the concentration of the CN -^ at each of the following volumes of added KCN:
Hg 2 2+^ will precipitate first and the equivalence point is at 20.00 mL. And the second equivalence point is at 30 mL. At 5 and 15 mL there is an excess of unreacted Hg 2 2+^. a. 5.00 mL (5points)
b. 15.00 mL (5points)
c. 35.00 mL (5points) At 35.00 mL, there are 5 mL excess of the [CN -^ ]:
11. Calculate the concentration of Ag+^ in saturated solutions of Ag 2 CO 3
(K sp = 8.1 10 -12^ ) in:
Supplemental information
