my favourite pet is the mole!

H1 Chemistry

(posted on 7 July 2008)

These are the explanations for the various MCQs in the recent Mid-year. Only questions with many answering incorrectly are presented as follow. For the rest of the questions not presented, please see Mr. Tong if you need help.

 

Question 1

The question asked for “electrons“, not “atoms of C”!

Number of mole of C = 4.0 / 12.0 = 0.3333 mol

One C atom has 6 electrons.

Number of electrons in 0.3333 mol of C = 0.3333 x 6 x 6.02 x 10²³ = 1.20 x 10²⁴ (Answer: C)

 

Question 3

Assume mass, in g, of SO₃ is x. Hence, the mass of NO₂ is 20-x.

Number of mole of SO₃ = x/80.1

Number of mole of NO₂ = (20-x)/46.0

1H₂SO₄ ≡ 2H⁺ and 1HNO₃ ≡ 1H⁺.

Number of mole of H⁺ = 0.450 = 2x/80.1 + 0.5(20-x)/46.0

Solving for x, x = 16.5 g

Hence, percentage of SO₃ = 82.5% (Answer: C)

 

Question 4

Write out balanced equation for all 4 compounds, the one which needs 4 mol of oxygen for every mole of organic compound burnt is CH₃COCH₃.

CH₃COCH₃ + 4O₂ → 3CO₂ + 3H₂O (Answer: C)

 

Question 5

Since 3 mol of Cl^- is oxidised, 3 mol of e^- is given out. (2Cl^- → Cl₂ + 2e^-)

This means that, 3 mol of e^- must have been accepted by 1 mol of MnO₄^-.

Oxidation number of Mn in MnO₄^- = +7 (Remember that this will go down by 3.)

Final oxidation number of Mn in product = +7 – 3 = +4

Oxidation number of Mn: in Mn²⁺ is +2, in Mn³⁺ is +3, in MnO₂ is +4, in MnO₄^2- is +6.

Answer C: MnO₂ where oxidation number of Mn is +4.

 

Question 6

angle of deflection α |charge/mass|

|charge/mass| of Mn²⁺ = 2/54.9 = 0.03643

|charge/mass| of Se^2- = 2/79.0 = 0.02532

Let x be the angle of deflection of Se^2-.

0.03643/0.02532 = 6/x

x = 4.17^o ≈ 4^o, to the nearest degree (Answer: B)

 

Question 7

Dissappointing for those who got it wrong. It’s a very simple question, as long as you remember these 2 things:

1. Electronic configuration for the neutral Cu atom must be drawn first, before that of Cu⁺ ion is obtained. You cannot calculate the number of electron first and then write the configuration!
2. Cu and Cr has slightly different manner of writing their electronic configuration. These two are the exceptions in the syllabus!

Cu atom: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s¹

Cu⁺ ion: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰

Hence, there is no half-filled orbital. (Answer: D)

 

Question 9

|LE| α |(q₊ x q_-)/(r₊ + r_-)|

You have to compare the 4 parameters one by one, giving priority to the charge rather than the radius, if need be. You cannot substitute the actual values into the formula to get a number for comparison!

Mg²⁺ has a higher cationic charge than Na⁺. Hence, option C and D are eliminated.

O^2- has smaller radius than S^2-. (Answer: A, since MgO has the largest magnitude of lattice energy.)

 

Question 12

The shapes of the various compounds given are, stated in order of them appearing in the various options:

A: trigonal planar, trigonal pyramidal, trigonal pyramidal

B: linear, bent, bent

C: all tetrahedral (Answer: C, since all the species have the same shape.)

D: linear, bent, bent

 

Question 13

Number of half-life cycle needed = 3

(1st from 0.080 to 0.040, 2nd from 0.040 to 0.020, 3rd from 0.020 to 0.010)

Length of time for each half-life = 1 hr / 3 = 20 mins (Answer: B)

 

 

(posted on 23 March 2008)

Stoichiometry and Redox Assignment 1

O Level days are gone! Please present your answer in a logical, coherent and scientifically acceptable manner. Some general presentation style to be adopted:

1. Please use blue or black ink for all work submitted (assignments, tests or exams). Work in pencil is considered as rough work and will not be marked. Green ink for correction. 
2. Please write clear and proper statement in calculations. Don’t just throw the numbers around and expect the examiners to follow your trail of thoughts. (erm, or your copied work, which is not acceptable too!)
3. Don’t introduce new signs, symbols or notations without explaining them. Don’t misuse signs like “:” (for ratio) or “→” (for writing chemical equation). So, “mass of HCl : 0.300 g” is not acceptable. Nor is “mass of HCl → 0.300 g”.
4. Follow the guidelines on decimal places (dp) and significant figures (sf) religiously. To summarise again, final answer to 3 sf. Intermediate workings to 4 or 5 sf. All A_r, M_r or molar mass to 1 dp (including that in workings, intermediates and final answer).
5. Please spell out “amt” and vol.” in full.
6. “Amount” refers to “number of mole”. So, avoid “amount of HCl = 25.0 cm³” or “amount of FA1 = 0.300 g”. use the word “volume” or “mass” instead.
7. “mol = 0.300 mol” or “mol of HCl = 0.300 mol” doesn’t make sense. Write out complete statement. The word or units “mols” doesn’t exist. It is wrong. It should be “mol”, without the “s”.
8. All answer (whether in intermediate workings or final answers) should be accompanied with units, except for A_r or M_r.

Please reset your calculator to “scientific mode” before starting calculation. This is because the value sometimes gets too small and is truncated on the calculator’s screen. So, “0.000000000036″ would appear as “0.00000000003″ for a calculator displaying only 12 digits (which is the case for most calculators anyway). If you use the latter value for subsequent calculation or present it as the final answer, your answer is going to be terribly wrong, and it’s not because you don’t know how to do! (which is going to be very very sad, but i won’t cry for you because you’ve been warned!) Setting it to “scientific mode” would allow he value to be displayed as “3.6 x 10^-11“, which is correct and also easier to read.

 

 

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The difficult part for this assignment is with Question 5 to 8. Hence, we’ll not go through Question 1 to 4. Let’s highlight Question 5 to 8 briefly.

 

 

 

 

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Question 5

 

 

Volume of H₂ in mixture = 25.0 – 17.0 = 8.0 cm³ [This is because CO reacts to give the same volume of CO₂. Hence, 17.0 cm³ gives the initial volume of CO and CO₂. Write out balanced equation to convince yourself.]

Volume of O₂ reacted = 13.0 – (19.0 – 17.0) = 11.0 cm³

Volume of O₂ reacted with H₂ = 8.0 ÷ 2 = 4.0 cm³ [Again, write out balanced equation to see why it's halved.]

Volume of O₂ reacted with CO = 11.0 – 4.0 = 7.0 cm³

Volume of CO in mixture = 7.0 x 2 = 14.0 cm³

Volume of CO₂ in mixture = 25.0 – 14.0 – 8.0 = 3.0 cm³

Hence, the various percentage compositions for H₂, CO and CO₂ are 32.0%, 56.0% and 12.0% respectively.

 

 

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Question 6

 

 

The balanced equation is, 2MnO₄^- + 5H₂O₂ + 6H⁺ → 2Mn²⁺ + 5O₂ + 8H₂O

Amount of MnO₄^- reacted = 3.857 x 10^-4 mol

Since 5H₂O₂ ≡ 2MnO₄^- 

Amount of H₂O₂ in 25.0 cm³ of diluted solution reacted = 9.6425 x 10^-4 mol

Amount of H₂O₂ in 250 cm³ of diluted solution = 9.6425 x 10^-3 mol = Amount of H₂O₂ in 10.0 cm³ of concentrated solution

[H₂O₂] of original concentrated solution = 9.6425 x 10^-3 ÷ (10.0/1000) = 0.964 mol dm^-3 (to 3 sf)

 

 

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Question 7

 

 

The balance equation is, I₂ + SO₂ + 2H₂O → SO₄^2- + 2I^- + 4H⁺

Amount of I₂ in 16.40 cm³ = 1.64 x 10^-4 mol = Amount of SO₂ in 50 cm³ sample of wine, [since 1 I₂ ≡ 1 SO₂].

[SO₂] in wine sample = 1.64 x 10^-4 ÷ (50/1000) = 3.28 x 10^-3 mol dm^-3

 

 

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Question 8

 

 

The balanced equations are,MnO₄^- + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O, and

Cr₂O₇^2- + 6Fe²⁺ + 14H⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H₂O

Amount of MnO₄^- = 3.6 x 10^-3 mol

Since 6MnO₄^- ≡ 30 Fe²⁺ ≡ 5Cr₂O₇^2-,

Amount of Cr₂O₇^2- = (3.6 x 10^-3 / 6) x 5 = 3.0 x 10^-3 mol

Volume of Cr₂O₇^2- required = 3.0 x 10^-3 / 0.150 = 20.00 cm³

 

 

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