CHEM 1201 : Test 2
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Fall 2011 Chem 1201 Test 2 Page 1
Formulas Given on Test 2
H = 6.63 10-34 Js c = 2.9979 108 m/s RH = 2.18 10-18 J NA = 6.022 1023
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1. Which of the following solutions will form a precipitate when mixed with K2S?
(A) KClO4
(B) Mn(NO3)2
(C) NH4NO3
(D) CsI
(E) CaCl2
Ans.: B
2. The box shown in the figure is filled with water and the aqueous particles allowed to react.
An exchange reaction occurs and a solid precipitate is formed. How many ions of each type
remain dissolved in the aqueous phase after complete reaction occurs?
(A) 4 NO3- ions, 9 Cl- ions
(B) 4 NO3- ions, 3 Al3+ ions, 5 Cl- ions
(C) 10 Cl- ions, 4 Ag+ions
(D) 4 NO3- ions, 3 Al3+ ions, 9 Cl- ions
(E) 4 Al3+ ions, 12 Cl- ions
Ans.: B
Al3+
Cl-
Cl-
Cl-
Ag+
NO3
-
Ag+
NO3
-
Ag+
NO3
-
Fall 2011 Chem 1201 Test 2 Page 2
3. Which of the following compounds makes a solution with the weakest conductivity when
dissolved in water? Assume 0.1 M concentrations for each solution.
(A) Fe(NO3)2
(B) NaOH
(C) (NH4)2SO4
(D) HCL
(E) weak acid HF
Ans.: E
4. How many grams of Cr2(SO4)3 are contained in a sample of 25 mL of Cr2(SO4)3 if SO42-
concentration is 0.6701 M. Please note that the given concentration is for the ion not for
Cr2(SO4)3.
(A) 10.052 g
(B) 19.711 g
(C) 6.5705 g
(D) 2.1902 g
(E) 0.6701 g
Ans.: D
5. An exchange reaction is carried out using a solution of aqueous lead (II) nitrate. A precipitate
is formed by adding a solution containing excess sodium hydroxide. The mass of the precipitate
was found to be 4.764 grams. What was the concentration of the original lead nitrate solution if
its original volume was 75.66 mL?
(A) 3.7935 M
(B) 5.2206 10-4 M
(C) 2.6103 10-4 M
(D) 0.52206 M
(E) 0.26103 M
Ans.: E
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Fall 2011 Chem 1201 Test 2 Page 3
6. A 1.00 L solution of 0.038 M lead (II) nitrate precipitate is prepared.
A 0.015 M sodium chloride is added to the lead (II) nitrate solution.
What volume of sodium chloride would be necessary to precipitate all of the lead ion from
solution?
(A) 1.2667 L
(B) 5.4213 L
(C) 2.5333 L
(D) 5.0667 L
(E) 4.408 L
Ans.: D
7. A 15.0 mL sample of an unknown concentration of the base barium hydroxide requires 25.0
mL of 0.1407 M of the acid HCl to each the equivalence point in a titration. What was the
concentration of the original barium hydroxide solution? Hint: Write a balanced reaction before
you begin.
(A) 8.7938 10-2 M
(B) 1.0553 M
(C) 4.403 10-2 M
(D) 0.23497 M
(E) 0.11725 M
Ans.: E
8. What is the concentration of nitrate ions in a solution made by mixing 325.0 mL of 0.2 M
potassium nitrate and 15 mL of 0.21 M calcium nitrate?
(A) 0.20044
(B) 0.19118
(C) 162.61
(D) 1.8529 10-2
(E) 0.20971
Ans.: E