OPTED sample test errors/typos (Full List)

[pamin002]

opted sample test errors/typos
test download: https://www.ada.org/oat/oat_sample_test.pdf

I have compiled a list of errors on the OPTED sample test (found at the link above). Please let me know if have found any additional typos.

The key goes like this:

• Opted Faulty Answer
  
  
• Most Correct Answer

​

NATURAL SCIENCES

44. If the mass percent of oxygen in a nitrogenoxygen
compound is known, given the mass
percent of oxygen, all of the following are
needed to determine the molecular formula of
a nitrogen-oxygen compound EXCEPT one.
Which one is this EXCEPTION?
A. Atomic mass of nitrogen
B. Atomic mass of oxygen
C. Avogadro's number
D. Empirical formula
E. Molar mass of the compound

**Although the Empirical formula is NOT needed, neither is Avogadro's number. There is no logical reason why you would need Avogadro's number.
-> The empirical formula would be useful for calculating the molecular formula, HOWEVER you can just as well use the 'mass percent of oxygen' combined with the atomic masses of each to figure out the molecular formula.
-> You would not use Avogadro's number to calculate the molecular formula.



61. A 0.60 M solution is made by dissolving solid
compound X in water. After ten seconds, the
concentration of X is 0.40 M. All of the
following could account for these results
EXCEPT one. Which one is this
EXCEPTION?
A. Precipitation
B. Neutralization
C. Evaporation
D. Decomposition
E. Disproportionation

**The answer given, Precipitation, would actually result in the decrease of the concentration.
--> On the other hand, Evaporation results in a decrease in volume. Molarity is found by using the formula: M = mols/volume -- thus, a decrease in volume would INCREASE the concentration, making this the correct answer.



93. What is the hybridization of a nitrogen atom if
it forms two σ two π bonds?
A. sp
B. sp2
C. sp3
D. sp3d2

** sp3 hybridization results from sigma bonds, and does not contain pi bonds.--> sp hybridization is a result of 2 sigma and 2 pi bonds --> making A the right answer.

PHYSICS

35. A 12-volt battery with an internal resistance of
1-Ω resistor. Which of the following is the
current in amperes that flows in the circuit.
A. 4
B. 3
C. 9
D. 12
E. 48

** 3 doesn't even make sense as an answer.
--> The current is found by using the formula: i = V/R. Plug in 12 as the voltage (V) and 1 as the resistance (R) --> i = 12/1 = 12.




(NOTE: I HAVE NOT CONFIRMED THIS LAST PROBLEM (PHYSICS # 39, please let me know if you disagree!)

39. What is the focal length in cm of a lens that
produces an image 30 cm behind it when the
object is placed 6 cm in front of it?
A. 7.5
B. 36
C. 5.0
D. 24
E. 18.0
**The focal length is found by: 1/f = 1/o + 1/i. Since the image is behind the lens, i will have a negative value (i = -30). Since the object is place in front of the lens, o will have a positive value (o = 7.5). SO, 1/f = 1/-30 + 1/6. Solve for f, and you get 7.5.



There you have it! I have confirmed the other answers with professors at my university (except physics #39) . If you disagree w/ anything, please indicate the number and why.

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[Thinh]

If they made a mistake on the sample problem I wonder if they made any mistakes on the actual test.

 

[blysssful]

If they made a mistake on the sample problem I wonder if they made any mistakes on the actual test.

Probably.

 

[OpticalBlackOut]

Great job, very nice of you to take the time out to do that.
Too bad I took the test last summer, haha.

Maybe this can be made into a sticky?

 

[pamin002]

Great job, very nice of you to take the time out to do that.
Too bad I took the test last summer, haha.

Maybe this can be made into a sticky?

haha thanks, yea im taking my OAT this sunday and i was working this test out and was going crazy over why i could not get the answer given in the key. i couldnt find a compilation of typos of this thing, so i thought id save other people in my position some work

 

[vlad925]

I would like to add this to the original list of errors

85. Which of the following will undergo a free radical bromination most rapidly?
A. CH3CH(CH3)CH3
B. CH3C(CH3)2CH3
C. CH4
D. CH3CH3
E. cyclopropane

**The answer given, CH4, is true for chlorination, since it has the most primary hydrogens per carbon, but is not the case for bromination.
--> Free radical bromination is 10^3 more reactive with tertiary hydrogens than with primary hydrogens. So CH3CH(CH3)CH3 must react more rapidly than CH4.

86. How many π (pi) molecular orbital does pyridine possess?
(picture of pyridine)
A. 3
B. 4
C. 5
D. 6
E. 7

**The answer given, 7, doesn't make sense.
--> There are 3 pi bonds and 6 pi electrons according to Hukels rule (4n+2), since the lone electrons in Nitrogen are not participating in the pi system. Technically a pi bond and p orbital is the same thing, so despite the strange wording of the question "pi orbital", the answer can only be 3.

Number 39. Physics question on Optics I believe the original answer is correct. Since the image is on the other side of the lens from the object, the image distance is going to be a positive number. i=30. When image and object are on the same side of the lens, "i" is negative. 1/f = 1/30 + 1/6 The answer works out to be C. 5cm

 

[quackquack]

86. How many π (pi) molecular orbital does pyridine possess?
(picture of pyridine)
A. 3
B. 4
C. 5
D. 6
E. 7

**The answer given, 7, doesn't make sense.
--> There are 3 pi bonds and 6 pi electrons according to Hukels rule (4n+2), since the lone electrons in Nitrogen are not participating in the pi system. Technically a pi bond and p orbital is the same thing, so despite the strange wording of the question "pi orbital", the answer can only be 3.

I'm not sure about this but... I think one PI bond = overlap of two pi orbitals. So there are 3 double bonds = 6 pi orbitals. The pi orbital for nitrogen is not part of the system, but the answer just says how many pi orbitals in total it has, so it would be 6 + 1 = 7. I think.

Also, I agree with you for the physics question!

 

[thecgrblue]

P-orbitals and Pi bonds are not the same at all. Also, why do you mention Hukles Rule? If the structure is aromatic or not has nothing to do with the question.

7 is the right answer.

 

[vandal]

7 is not the answer. The correct answer is 6. The author of the answer key might have thought that lone pair on the nitrogen were part of a pi orbital, but it is not. It's in sp2.

I asked my organic chem professor to confirm. Hope this clears things up.

 

[euphaire]

Is the solution to Q87 really B?
To synthesize a tertiary alcohol, grinard reagents must be used. I don't know any other way. The solution says a ketone plus KOH and KCN. I know KOH will deprotonate the ketone on the alpha carbon, so you get an enolate ion. Then what? How can you end up getting a tertiary alcohol from there?

Thanks in advance

 

[euphaire]

I talked to an organic chemistry professor at UW today regarding question 87 and 89. Surprisingly, he told me those were TERRIBLE questions because they were simply...wrong. I'd like to share with you guys what he said about them.

87. Which combination of reagents will produce
None of the given answers were right! The solution says it's B, but more reagent is needed. For instance, KCN attacks carbonyl carbon, then OH hydrolyzes CN, turning it into a COOH group, from there you need Wolf-Kishner or Clemmensen which means more reagents are needed.

89.
It's a bad question because Imidazole isn't acidic to start with. Imidazole itself is already aromatic. After removing the H, the ion is still aromatic. There's no change in aromaticity after deprotonating, so that's why asking what's causing the conjugate base to be stable is a bad idea. Anyway, the solution they want you to look at is D, which is "correct", but its' just a terrible question and a terrible way of asking it.

I hope this helped people who were as confused as I was. I trust the professor, so I'm pretty sure he's right about this...

Cherry

 

[nucci96]

I would like to add this to the original list of errors

85. Which of the following will undergo a free radical bromination most rapidly?
A. CH3CH(CH3)CH3
B. CH3C(CH3)2CH3
C. CH4
D. CH3CH3
E. cyclopropane

**The answer given, CH4, is true for chlorination, since it has the most primary hydrogens per carbon, but is not the case for bromination.
--> Free radical bromination is 10^3 more reactive with tertiary hydrogens than with primary hydrogens. So CH3CH(CH3)CH3 must react more rapidly than CH4.

86. How many π (pi) molecular orbital does pyridine possess?
(picture of pyridine)
A. 3
B. 4
C. 5
D. 6
E. 7

**The answer given, 7, doesn't make sense.
--> There are 3 pi bonds and 6 pi electrons according to Hukels rule (4n+2), since the lone electrons in Nitrogen are not participating in the pi system. Technically a pi bond and p orbital is the same thing, so despite the strange wording of the question "pi orbital", the answer can only be 3.

Number 39. Physics question on Optics I believe the original answer is correct. Since the image is on the other side of the lens from the object, the image distance is going to be a positive number. i=30. When image and object are on the same side of the lens, "i" is negative. 1/f = 1/30 + 1/6 The answer works out to be C. 5cm

Is their answer for #85 definitely wrong?