Multiple-choice (MC) exam tips

Success on a MC exam requires both subject matter knowledge AND test taking skills.  A student with good test taking skills with average subject matter knowledge could have a significantly higher score than a student with poor test taking skills with good subject matter knowledge, i.e. (unfortunately ?) good test taking skills are valuable. 

To improve MC exam test taking skills:  practice, practice, & practice - using e.g.  AP Chemistry study guides, SAT II Chemistry study guides, & California High School Chemistry exam study guides.

• if there are 2 choices with opposite or similar meaning, then one of the choices may be the answer

• work quickly:  do the easy questions and skip the hard questions
• monitor your time; pace yourself during the exam [bring a watch]

  you will have about 1 minute / question (on the AP or SAT II Chemistry exam), so don't watse your time on hard questions - you earn the same number of points for an easy question as a difficult question, but doing the hard question may prevent you from doing the easy questions - the net effect is getting a lower test score. 

  first, answer the easy questions, skip questions that you do not know, and mark questions that you can do, but would take more than a minute, then after completing the easy questions, return to these questions that would take longer than a minute.

• exam format:  penalty for wrong answer

  eliminate any wrong answers; if you can eliminate at least one wrong answer, then guess

  do not guess, when you can not eliminate any choice

• marking exam questions

  identify:  [useful when you return to that question - to either try again or to check it ]

  wrong answers by a slash

  uncertain choices by a question mark

  circle the correct answer

  do not mark your answer sheet for each question; rather do it for a group of questions - more efficient use of time

• read questions / answers carefully [note: key words would NOT be high-lighted in the exam]

  ex1.  Which aqueous solution has the highest freezing point ?

  a.  100 mL 2.0 M glucose
  b.  100 mL 1.0 M sodium chloride
  c.  300 mL 0.5 M sucrose
  d.  50 mL 3.0 M sucrose
  e.  2 or more of the above solutions would have the highest freezing point.

  analysis:   the equation, Δ T_f = i K_f m, refers to freezing point depression, so the highest freezing point would be the solution with the smallest Δ T_f , hence the smallest product of "im"  [concentration of particles in solution;  the volume is irrelevant]. 

  the answer would not be choice d, since it has the highest [particles]. 

  while choice a & b (apparently) have the same [particles], thus you might pick choice e as your answer, it would be wrong.  in choice a, [particles due to glucose] = 1 *   2 M = 2 M, while for choice b, [particles due to NaCl] ≈ 2 * 1 M = 2 M.  the factor, i, for NaCl is about 2 - its actual value is slightly less than 2 (i.e. the dissociation of NaCl is not "perfect"); thus [particles due to NaCl] < 2 M, hence the answer = choice b.

  ex2.  The vapor pressure of a solution of ethanol (CH₃CH₂OH) in an open container will

  a.  increase over  time
  b.  decrease over time
  c.  be a constant over time
  d.  be zero
  e.  oscillate over  time

  analysis:  the term "open" in the question means that ethanol would evaporate and eventually be completely gone, so choice b = answer.   on the other hand, if it was a closed container, then the vapor pressure would be a constant.

  ex3.  An unique characteristic of the alkaline earth metals is

  a.  it has properties of metals
  b.  its atomic size increases as its atomic number increases
  c.  its ionization energy tends to decrease as its atomic number increases
  d.  its cation has a charge of 2
  e.  all of the above

  analysis:  the term, unique, means that it is not found among other families of elements.  while all of the choices are true, the answer is not e, since only the alkaline earth metals, as a family, has its cation with a charge of 2, i.e. answer = choice d.

  ex4.  A 100 mL 0.5 M magnesium hydroxide solution would be neutralized by a solution of

  a.  100 mL 0.5 M HCl
  b.  100 mL 0.5 M KOH
  c.  100 mL 0.5 M NaCl
  d.  50 mL 1 M H₂SO₄
  e.  200 mL 0.5 M HNO₃

  analysis:  magnesium hydroxide is Mg(OH)₂; thus answer = choice e.  notice that the equation to solve acid - base titration problems, a [acid] V_acid = b [base] V_base, has terms, a & b, which refers to the # of protons or hydroxide ions formed from the acid or base, respectively.

   

• Roman numerial questions : determine if a statement is true or false, then eliminate choice(s) accordingly

  ex.   How do isomers differ ?

  I.  bonds
  II.  chemical formula
  III.  physical properties
  IV.  arrangment of atoms in space

  a.  III only
  b.  II and IV
  c.  II, III, and IV
  d.  I, III, and IV
  e.  I, II, III, and IV

  analysis:        by knowing that IV is true, you can eliminate choice a

  by knowing that II is false, you can eliminate choices b, c, & e

  therefore, answer = choice d

• calculations:  calculators are not allowed for MC questions

usually, the choices are quite different, so you need to only estimate the answer - do not calculate the actual answer - it could be a waste of your time.  also, use factoring.

ex1.  % Ag in Ag(NH₄)₂ is ___.

a.  4.2
b.  19.7
c.  76.1
d.  80.3
e.  95.8

analysis:   answer     =    mass of Ag / mass of Ag(NH₄)₂

=   108 / (108 + 28 + 8)   =   108 / 144 

=  54 / 72  =  27 / 36  =  3 / 4

therefore, answer = choice c

 

ex2.  As_xO_y contains 65.2% As.  What is its empirical formula ?

a.  AsO
b.  As₂O₃
c.  AsO₂
d.  As₂O₅
e.  As₂O

analysis:   look-up the atomic masses of As & O:   As = 75  &  O = 16

eliminate choice a (AsO);     As : O is

75 : 16  -->  % As = 75 / (75+16) = 75 / 91 ≈ 75 / 100 = 75% ≠ 65.2%

eliminate choice b (As₂O₃);  As : O is

150 : 48 --> % As = 150 /(150+48) = 150 / 198 ≈ 150 / 200 = 75% ≠ 65.2%

eliminate choice c (AsO₂);   As : O is

75 : 32  -->  % As = 75 / (75+32) = 75 / 107 ≈ 75 / 100 = 75% ≠ 65.2%

choice d (As₂O₅):    As : O is

150 : 80 -->  % As = 150 / (150+80) = 150 / 230 = 15 / 23 ≈ 15 / 25 = 3 / 5 = 60%  -->  possible answer.

eliminate choice e  (As₂O);   As : O is

150 : 16 -->  % As = 150 / (150+16) = 150 / 166 >> 50% ≠ 65.2%

alternatively  150 / 166 = 90% ≠ 65.2%,

since 150 = 100% (166) - 10% (166) = 90% (166)

thus 150 / 166 = 90 %

ex3.   . . .

answer = 250 * 0.8 * 2.4 * 22  =  [(25 * 10) * (4 * 2 * 0.1)] * 2.4 * 22 

=  200 * 2.4 * 22  ≈  200 * 2.5 * 20 = (2*100) * 2.5 * (2*10)

= 2 * 5 * 1,000 = 10,000

 

• SAT 2 Chemistry:  T / F;  statement I / statement II format questions

  there must be at least 2 marks per question; otherwise, your answer will be wrong, i.e. do not mark only statement I or II as true or false, then ignore the other statement

  do not mark CE if statement I or II is false;  both statements I and II must be true - to consider marking CE