IB Chemistry 화학 S-1 시험 출제 예상 범위 해설 (제 학생들만 볼 수 있습니다.)

Enter password to access the page:

Submit

Question 1: Magnesium Oxide Reaction and Yield Calculation

(a)(i) Balanced Equation for Magnesium Oxide Reaction

The balanced chemical equation for magnesium reacting with oxygen is:

2Mg(s) + O₂(g) → 2MgO(s)

(a)(ii) Metal in the Same Period as Magnesium that Does Not Form a Basic Oxide

Al - amphoteric

(b)(i) Calculate the Amount of Magnesium Used

Mass of magnesium used:

6.354 g (calculated by subtracting the mass of the crucible and lid from the mass before heating).

The molar mass of magnesium (Mg) is 24.305 g/mol. Using this, the amount of magnesium in moles is:

Moles of magnesium = 0.2615 mol

(b)(ii) Determine the Percentage Uncertainty of the Mass of Product After Heating

The mass of the product is calculated as:

9.569 g

The total uncertainty is ±0.002 g. Therefore, the percentage uncertainty is:

Percentage uncertainty = (0.002 / 9.569) × 100 = 0.02%

(b)(iii) Calculate the Percentage Yield of Magnesium Oxide

Theoretical mass of magnesium oxide (MgO) based on 0.2615 mol of magnesium:

Theoretical mass = 10.5 g

Actual mass of magnesium oxide (MgO) obtained:

Actual mass = 9.6 g

Therefore, the percentage yield is:

Percentage yield = (9.6 / 10.5) × 100 = 91%

(b)(iv) Could Magnesium Reacting with Nitrogen Explain the Yield?

If some magnesium reacted with nitrogen to form magnesium nitride (Mg₃N₂), it would result in a lower yield of MgO. Thus, this reaction could explain the lower yield.

(b)(v) Other Possible Explanation for Yield Reduction

Another possible explanation for the reduced yield could be incomplete reaction of the magnesium, where not all of it reacted to form MgO.

Question 2: Lithium Reaction with Water

(a) Balance the Equation for the Reaction of Lithium with Water

The balanced equation for lithium reacting with water is:

2Li(s) + 2H₂O(l) → 2LiOH(aq) + H₂(g)

(b) Calculate the Molar Concentration of Lithium Hydroxide

Given:

• Mass of lithium: 0.200 g
• Volume of water: 500.0 cm³ = 0.500 dm³
• Molar mass of lithium (Li): 6.94 g/mol

Calculate the moles of lithium:

Moles of Li = 0.200 g / 6.94 g/mol = 0.02883 mol

Since the ratio of Li to LiOH in the balanced equation is 1:1, the moles of lithium hydroxide formed is also 0.02883 mol.

Calculate the molar concentration of lithium hydroxide:

Concentration of LiOH = 0.02883 mol / 0.500 dm³ = 0.05766 mol/dm³

(c) Calculate the Volume of Hydrogen Gas Produced

Given:

• Temperature = 25°C = 298 K
• Pressure = 101 kPa
• Moles of H₂ = 0.014415 mol (half the moles of Li since 1 mol of H₂ is produced per 2 mol of Li)
• Ideal gas constant, R = 8.31 J/mol·K

Use the ideal gas equation:

PV = nRT

Solve for volume V:

V = (nRT) / P = (0.014415 mol × 8.31 J/mol·K × 298 K) / 101 = 0.354 dm³ = 354 cm³

(d) Suggest a Reason Why the Volume of Hydrogen Gas Collected Was Smaller Than Predicted

A possible reason for the smaller volume of hydrogen gas could be loss of gas during the collection process, such as gas escaping before being fully collected or dissolving into the water.

Question 3: Calcium Carbonate in Eggshell

(a) Calculate the Amount of HCl Added

Given:

• Volume of HCl = 27.20 cm³ = 0.02720 dm³
• Concentration of HCl = 0.200 mol/dm³

Calculate the moles of HCl added:

Moles of HCl = Volume × Concentration = 0.02720 dm³ × 0.200 mol/dm³ = 0.00544 mol

(b) Calculate the Amount of Excess HCl

Given:

• Volume of NaOH used = 23.80 cm³ = 0.02380 dm³
• Concentration of NaOH = 0.100 mol/dm³

Calculate the moles of NaOH used to neutralize the excess HCl:

Moles of NaOH = Volume × Concentration = 0.02380 dm³ × 0.100 mol/dm³ = 0.00238 mol

Since HCl and NaOH react in a 1:1 ratio, the moles of excess HCl are also 0.00238 mol.

(c) Determine the Amount of HCl that Reacted with Calcium Carbonate

The total amount of HCl added was 0.00544 mol, and the excess HCl was 0.00238 mol. Therefore, the amount of HCl that reacted with calcium carbonate is:

Moles of HCl that reacted = 0.00544 mol − 0.00238 mol = 0.00306 mol

(d) State the Equation for the Reaction of HCl with Calcium Carbonate

The balanced chemical equation is:

CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + CO₂(g) + H₂O(l)

(e) Determine the Amount of Calcium Carbonate in the Eggshell

From the balanced equation, 1 mol of CaCO₃ reacts with 2 mol of HCl. Therefore, the moles of CaCO₃ that reacted are:

Moles of CaCO₃ = 0.00306 mol HCl / 2 = 0.00153 mol

(f) Calculate the Mass and Percentage by Mass of Calcium Carbonate in the Eggshell Sample

Molar mass of CaCO₃ = 100.09 g/mol

Mass of CaCO₃ = 0.00153 mol × 100.09 g/mol = 0.153 g

Percentage by mass of CaCO₃ in the eggshell:

Percentage by mass = (0.153 g / 0.188 g) × 100 = 81.4%

(g) Deduce One Assumption Made in Arriving at the Percentage of Calcium Carbonate in the Eggshell

One assumption made is that all of the calcium carbonate in the eggshell reacted completely with the hydrochloric acid, meaning no unreacted CaCO₃ remained in the sample.

Question 4: Magnesium and Hydrochloric Acid Reaction

(a) Balanced Equation for Magnesium and Hydrochloric Acid Reaction

The balanced equation for magnesium reacting with hydrochloric acid is:

Mg(s) + 2HCl(aq) → H₂(g) + MgCl₂(aq)

(b)(i) Moles of Hydrogen Gas (H₂) Produced

Given:

• Mass of magnesium (Mg): 0.0740 g
• Molar mass of magnesium (Mg): 24.31 g/mol

Calculate the moles of magnesium:

n(Mg) = 0.0740 g / 24.31 g/mol = 3.04 × 10⁻³ mol

Hydrochloric acid (HCl) concentration and volume:

• Concentration of HCl: 2.00 mol/dm³
• Volume of HCl: 15.0 cm³ = 0.0150 dm³

Calculate the moles of HCl:

n(HCl) = 2.00 mol/dm³ × 0.0150 dm³ = 3.00 × 10⁻² mol

(b)(ii) Theoretical Yield of Hydrogen Gas (H₂) in Moles

Since the reaction produces 1 mol of H₂ per mol of Mg:

n(H₂) = n(Mg) = 3.04 × 10⁻³ mol

(c) Theoretical Volume of Hydrogen Gas (H₂) at the Given Conditions

Use the ideal gas equation:

V = (nRT) / P

Where:

• n = 3.04 × 10⁻³ mol
• R = 8.31 J/mol·K
• T = 293 K (20.0 °C)
• P = 1.01 × 10⁵ Pa

Calculate the volume:

V = (3.04 × 10⁻³ mol × 8.31 × 293) / (1.01 × 10⁵) = 73.4 cm³

(d) Reasons for the Lower Actual Volume of Hydrogen Gas

The actual volume of hydrogen gas measured may be lower due to:

• Gas leaks from the apparatus
• Syringe sticking during the collection
• Impurities in the magnesium

Question 5: Menthol Combustion and Molar Mass Calculation

(a) Determine the Empirical Formula of Menthol

Given:

• Mass of menthol (C, H, O): 0.1595 g
• Mass of CO₂ produced: 0.4490 g
• Mass of H₂O produced: 0.1840 g

Step 1: Calculate the moles of carbon (C):

Moles of C = 0.4490 g CO₂ / 44.01 g/mol = 0.01020 mol

Step 2: Calculate the moles of hydrogen (H):

Moles of H = (0.1840 g H₂O × 2) / 18.02 g/mol = 0.02042 mol

Step 3: Calculate the moles of oxygen (O):

Total mass of menthol = 0.1595 g

Mass of C and H = 0.1225 g (C) + 0.0206 g (H) = 0.1431 g

Mass of oxygen (O) = 0.1595 g − 0.1431 g = 0.0164 g

Moles of O = 0.0164 g / 16.00 g/mol = 0.001025 mol

Step 4: Empirical formula:

C₁₀H₂₀O

Award [3] for correct final answer.

(b) Calculate the Molar Mass of Menthol

Given:

• Mass of menthol sample: 0.150 g
• Volume of vaporized menthol: 0.0337 dm³
• Temperature: 150 °C = 423 K
• Pressure: 100.2 kPa = 100,200 Pa
• Gas constant (R) = 8.31 J/mol·K

Use the ideal gas equation:

pV = nRT and n = m/M

Solve for M (molar mass):

M = (0.150 g × 8.31 J/mol·K × 423 K) / (100.2 kPa × 0.0337 dm³)

Calculate the molar mass:

M = 156 g/mol

Award [1] for correct answer with no working shown.

Question 6: Determining Empirical and Molecular Formula of HX

(a) Determine the Empirical Formula of HX

Given the mass composition:

• C = 39.99%
• H = 6.73%
• O = 53.28%

Step 1: Assume a 100 g sample, so we have:

• 39.99 g of C
• 6.73 g of H
• 53.28 g of O

Step 2: Calculate the moles of each element:

Moles of C = 39.99 g / 12.01 g/mol = 3.33 mol

Moles of H = 6.73 g / 1.008 g/mol = 6.68 mol

Moles of O = 53.28 g / 16.00 g/mol = 3.33 mol

Step 3: Simplify the mole ratio:

• C:H:O = 3.33 : 6.68 : 3.33
• Dividing by the smallest value (3.33), we get C:H:O = 1:2:1

Empirical formula of HX: CH₂O

(b) Calculate the Molar Mass of HX

Given:

• Mass of HX = 1.51 g
• Volume of HX solution = 25.00 cm³
• Volume of NaOH solution used = 22.30 cm³
• Concentration of NaOH = 0.750 mol/dm³

Step 1: Calculate the moles of NaOH:

n(NaOH) = concentration × volume = 0.750 mol/dm³ × 0.02230 dm³ = 0.016725 mol

Step 2: Since HX is monoprotic, moles of HX = moles of NaOH:

n(HX) = 0.016725 mol

Step 3: Calculate the molar mass of HX:

Molar mass = mass / moles = 1.51 g / 0.016725 mol = 90.3 g/mol

(c) State the Molecular Formula of HX

The empirical formula mass of CH₂O is:

12.01 (C) + 2 × 1.008 (H) + 16.00 (O) = 30.03 g/mol

Molecular mass / empirical formula mass = 90.3 g/mol / 30.03 g/mol ≈ 3

Therefore, the molecular formula is:

C₃H₆O₃

(d) Identify a Functional Group in HX

The reaction of HX with NaOH suggests that HX contains a carboxyl group (-COOH).

Question 7: Double Salts and Hydrated Sulfate

(a) Composition of the Double Salt

Given the percentage composition of the double salt:

Element	Percentage (%)
Nitrogen (N)	7.09%
Hydrogen (H)	5.11%
Sulfur (S)	16.22%
Cobalt (Co)	14.91%
Oxygen (O)	---

(b) Lewis Structure of Sulfate Ion

The Lewis structure of the sulfate ion (SO₄²⁻) can be represented as:

Sulfur is in the center with four oxygen atoms surrounding it, and each oxygen atom has a double bond with sulfur. There are two extra electrons, giving the ion a 2− charge.

(c) Calculate the Percentage of Oxygen

To calculate the percentage of oxygen, we first need to subtract the total percentage of other elements from 100%:

• Total percentage of N, H, S, and Co = 7.09% + 5.11% + 16.22% + 14.91% = 43.33%
• Percentage of oxygen = 100% − 43.33% = 56.67%

(d) Determine the Empirical Formula

Using the percentage composition and the molar masses of each element, we can determine the empirical formula:

• Nitrogen (N): 7.09 g / 14.01 g/mol = 0.506 mol
• Hydrogen (H): 5.11 g / 1.008 g/mol = 5.07 mol
• Sulfur (S): 16.22 g / 32.07 g/mol = 0.506 mol
• Cobalt (Co): 14.91 g / 58.93 g/mol = 0.253 mol
• Oxygen (O): 56.67 g / 16.00 g/mol = 3.54 mol

Step 1: Simplify Mole Ratios

To simplify the mole ratios, we divide all moles by the smallest mole value, which is for Cobalt (0.253 mol):

• Nitrogen (N): 0.506 mol / 0.253 mol ≈ 2
• Hydrogen (H): 5.07 mol / 0.253 mol ≈ 20
• Sulfur (S): 0.506 mol / 0.253 mol ≈ 2
• Cobalt (Co): 0.253 mol / 0.253 mol = 1
• Oxygen (O): 3.54 mol / 0.253 mol ≈ 14

Step 2: Molecular Formula

Based on the simplified mole ratios, the empirical formula of the double salt is:

Co(NH₄)₂(SO₄)₂ · 6H₂O

(e) Ionic Equation for the Reaction with Barium Chloride

The ionic equation for the precipitation of sulfate ions as barium sulfate is:

SO₄²⁻(aq) + Ba²⁺(aq) → BaSO₄(s)

(f) Calculate the Mass of Barium Sulfate Precipitate

Using the empirical formula and molar mass of the double salt Co(NH₄)₂(SO₄)₂ · 6H₂O, we calculated the following:

• Molar mass of the double salt = 395.27 g/mol
• Moles of the double salt = 1.20 g / 395.27 g/mol = 0.00304 mol
• Moles of sulfate ions = 0.00304 mol × 2 = 0.00607 mol
• The molar mass of BaSO₄ is 233.39 g/mol
• Mass of BaSO₄ precipitate = 0.00607 mol × 233.39 g/mol = 1.42 g

Question 8: White Phosphorus and Phosphine Production

(a) Calculate the Amount of White Phosphorus Used

Given:

• Mass of white phosphorus (P₄) = 2.478 g
• Molar mass of P₄ = 4 × 30.97 g/mol = 123.88 g/mol

Moles of P₄ = 2.478 g / 123.88 g/mol = 0.0200 mol

(b) Determine the Limiting Reagent

Given:

• Volume of NaOH = 100.0 cm³ = 0.100 dm³
• Concentration of NaOH = 5.00 mol/dm³

Moles of NaOH = 5.00 mol/dm³ × 0.100 dm³ = 0.500 mol

From the equation, 3 mol of NaOH reacts with 1 mol of P₄. Therefore, 0.0200 mol of P₄ requires 0.0600 mol of NaOH, which is much less than the available NaOH (0.500 mol). Thus, P₄ is the limiting reagent.

(c) Determine the Excess Amount of NaOH

Excess NaOH = 0.500 mol − 0.0600 mol = 0.440 mol

(d) Determine the Volume of Phosphine (PH₃) Produced

The balanced equation shows that 1 mol of P₄ produces 1 mol of PH₃. Therefore, the moles of PH₃ produced = moles of P₄ used = 0.0200 mol.

At standard temperature and pressure (STP), 1 mol of gas now occupies 22.7 dm³. Therefore:

Volume of PH₃ = 0.0200 mol × 22.7 dm³/mol = 0.454 dm³ = 454 cm³

Question 9: Titration and Percentage Uncertainty

(a) Calculate the Percentage Uncertainty Associated with the Titre

Given:

• Titre value = 3.1 cm³
• Uncertainty = ±0.1 cm³

Percentage uncertainty = (0.1 cm³ / 3.1 cm³) × 100 = 3.23%

(b) Suggest One Procedural Modification to Reduce the Percentage Uncertainty

One way to reduce the percentage uncertainty is to increase the volume of titrant used. This can be achieved by using a more concentrated analyte or increasing the sample size.

(c) Determine the Percentage of Iron in the Kale Leaves

Problem cannot be solved due to insufficient information.

(d) Suggest One Reason for the Large Discrepancy in Iron Content

One possible reason for the discrepancy could be contamination of the sample, which could artificially inflate the iron content.

----------------------------------------------------------------------------------------------------------

Question 1

• First ionization energy of calcium vs potassium: Calcium has a higher ionization energy because it has more protons, which increases nuclear attraction on the outer electrons.
• Limitations of electron energy level model: This model does not apply to hydrogen's subenergy levels or account for electron-electron repulsion in multi-electron atoms.
• Arrow X: From n=1 to infinity (ionization of ground-state hydrogen).
• Arrow Z: From n=3 to n=2 (lowest energy transition in the visible spectrum).

Question 2

• Atoms are no longer indivisible: Discovery of subatomic particles and nuclear reactions.
• Bonding in magnesium, oxygen, and magnesium oxide:

  Substance	Bond Type	Valence Electron Behavior
  Magnesium	Metallic	Delocalized electrons form a 'sea' of electrons.
  Oxygen	Covalent	Shared electrons form a covalent bond.
  Magnesium Oxide (MgO)	Ionic	Electrons are transferred from Mg to O, forming ions.

Question 3

• 14N³⁻ Subatomic Particles: Protons: 7, Neutrons: 7, Electrons: 10
• Relationship between 14N³⁻ and 15N³⁻: Isotopes (same number of protons, different neutrons).
• Greater ionic radius: N³⁻ has a greater ionic radius due to more electron-electron repulsion.
• Sixth ionization energy: Nitrogen has a higher sixth ionization energy because of its half-filled p-orbitals.

Question 4

• Greater ionization energy: Cu²⁺ has a higher ionization energy because it has fewer electrons, increasing nuclear attraction on the remaining electrons.
• Frequency for ionization of copper: Approximately \(1.17 \times 10^{15}\) s⁻¹.
• Magnetic properties of iron: Iron has unpaired electrons in the d-orbitals, making it paramagnetic.

Question 5

• Fe²⁺ electron configuration: [Ar] 3d⁶
• Relative atomic mass of copper: Approximately 63.62, calculated from isotope abundances.

Question 6

• Ground-state orbital diagram of boron: 1s² 2s² 2p¹
• Shapes of orbitals: 1s (spherical), 2s (spherical), 2p (dumbbell-shaped).

Question 7

• Minimum frequency to break Cl₂ bond: Approximately \(6.09 \times 10^{14}\) s⁻¹.

Question 8

• Technique to determine argon isotope proportions: Mass spectrometry.

Question 9

• Conclusion from gold foil experiment: Most alpha particles passed through, indicating atoms are mostly empty space. A few were deflected, showing a small, dense nucleus.
• Electron configuration of copper: [Ar] 3d¹⁰ 4s¹
• Color change in copper complex: The d-orbitals are split, and electrons are promoted between the split orbitals, absorbing light in the visible region and emitting a complementary color.