DP Chemistry (last assessment 2024)

Question 20N.3.hl.TZ0.12

Date	November 2020	Marks available	[Maximum mark: 10]	Reference code	20N.3.hl.TZ0.12
Level	hl	Paper	3	Time zone	TZ0
Command term	Calculate, Deduce, Determine, Outline, Show	Question number	12	Adapted from	N/A

12.

[Maximum mark: 10]

20N.3.hl.TZ0.12

$1.57 %$ of the mass of a rock weighing $46.5 kg$ is uranium(IV) oxide, ${UO}_{2}$ . $99.28 %$ of the uranium atoms in the rock are uranium-238, ${}^{238}U$ .

(a)

Show that the mass of the ²³⁸U isotope in the rock is $0.639 kg$ .

[2]

Markscheme

$« \frac{mass %}{fraction of U in {UO}_{2}} = » \frac{238.03}{238.03 + 2 \times 16}$ / $0.881$ / $88.1 %$ ✔

$46.5 « kg » \times 0.0157 \times 0.881 \times 0.9928 « = 0.639 kg »$ ✔

Award [1 max] for omitting mass composition (giving $0.725 k g$ ).

M2 is for numerical setup, not for final value of $0.639 k g$ .

Examiners report

This question was generally well answered. Many candidates approached this question using amount, in mol, and converting to mass at the end. Some candidates omitted the mass composition however, resulting in a mass of 0.725 kg, which yielded [1 max].

(b)

The half-life of ²³⁸U is $4.46 \times 10^{9}$ years. Calculate the mass of ²³⁸U that remains after $0.639 kg$ has decayed for $2.23 \times 10^{10}$ years.

[2]

Markscheme

Alternative 1
$« \frac{2.23 \times 10^{10} year}{4.46 \times 10^{9} year} = » 5.00 « half - lives »$ ✔

$« m = 0.639 kg \times (0.5)^{5} = » 0.0200 « kg »$ ✔

Alternative 2
$« 𝜆 = \frac{\ln 2}{4.46 \times 10^{9} year} = » 1.554 \times 10^{- 10} « {year}^{- 1} »$ ✔

$« m = 0.639 kg \times 𝑒^{- 1.554 \times 10^{- 10 {year}^{- 1}} \times 2.23 \times 10^{10 year}} = » 0.0200 « kg »$ ✔

Award [2] for correct final answer.

Examiners report

The question involving half-life was very well answered and most scored both marks giving a final answer of 0.0200 kg, via different methods of calculation.

(c)

Outline a health risk produced by exposure to radioactive decay.

[1]

Markscheme

Any one:

«genetic» mutations ✔

«could cause» cancer ✔
Accept specific named types of cancer.

cells «in body» altered ✔

cells «in body» cannot function ✔

damaged DNA/proteins/enzymes/organs/tissue ✔

«radiation» burns ✔

hair loss ✔

damage in foetuses ✔

damages/weakens immune system ✔

Examiners report

This question on outlining a health risk produced by exposure to radioactive decay posed no difficulty and the most common, correct answers included "could cause cancer" and "can damage DNA".

(d)

Deduce the nuclear equation for the decay of uranium-238 to thorium-234.

[1]

Markscheme

${}_{92}^{238}U \to {}_{90}^{234}{Th} + {}_{2}^{4}{He}$ ✔

Do not penalize missing atomic numbers in the equation.

Accept “ $α$ ” for " $H e$ ”.

Examiners report

Most scored the one mark for the nuclear equation for the decay of uranium-238 to thorium-234. The most common error was including a neutron on the product side

(e)

Thorium-234 has a higher binding energy per nucleon than uranium-238. Outline what is meant by the binding energy of a nucleus.

[1]

Markscheme

energy required to separate a nucleus into protons and neutrons/nucleons
OR
energy released when nucleus was formed from «individual/free/isolated» protons and neutrons/nucleons ✔

Do not accept “energy released when atom was formed”.

Examiners report

Although many scored the one mark for outlining what is meant by the binding energy of a nucleus, the question was often answered as energy involved in the formation of an atom. Some candidates did not understand the difference between nucleus and nucleons.

(f)

Determine the nuclear binding energy, in $J$ , of ${}_{238}U$ using sections 2 and 4 of the data booklet.

The mass of the ${}_{238}U$ nucleus is $238.050786 amu$ .

[3]

Markscheme

$238.050786 « amu » \times 1.66 \times 10^{- 27} « kg {amu}^{- 1} »$
OR

✔

$(92 \times 1.672622 \times 10^{- 27}) + (146 \times 1.674927 \times 10^{- 27}) - 3.95 \times 10^{- 25}$
OR
$3.42 \times 10^{- 27} / 3 \times 10^{- 27} « kg »$ ✔

$« E = m c^{2} = 3.42 \times 10^{- 27} \times (3.00 \times 10^{8})^{2} = » 3.08 \times 10^{- 10} « J »$ ✔

Accept answers in the range “ $2.7 \times 10^{- 10} - 3.1 \times 10^{- 10} « J »$ ”.

Award [3] for correct final answer.

Examiners report

This demanding question on the determination of the nuclear binding energy was well executed and many scored all three marks. Even the weaker candidates still managed to gain an ECF mark for M3 by using the E = mc² equation. The general performance on this type of calculation was much better than in previous sessions.

Syllabus sections

Core

Core » Topic 1: Stoichiometric relationships » 1.3 Reacting masses and volumes

Core » Topic 1: Stoichiometric relationships

Options

Options » C: Energy » C.3 Nuclear fusion and fission

Options » C: Energy

Options » C: Energy » C.7 Nuclear fusion and nuclear fission (HL only)

Question 20N.3.hl.TZ0.12

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