DP Chemistry (last assessment 2024)

Question 20N.3.sl.TZ0.10

Date	November 2020	Marks available	[Maximum mark: 7]	Reference code	20N.3.sl.TZ0.10
Level	sl	Paper	3	Time zone	TZ0
Command term	Calculate, Deduce, Outline, Show	Question number	10	Adapted from	N/A

10.

[Maximum mark: 7]

20N.3.sl.TZ0.10

$1.57 %$ 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 ${}_{238}U$ 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 was a very different question as student were given the answer and it was the work that was being marked. Students should always clearly show their calculations so examiners can award marks throughout the question and potentially award ECF if possible. It is very difficult to do this when students do not show work clearly.

(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

This continues to be a topic that students understand well, and probably more used alternative 1 to calculate the half-life. Students should always clearly show their calculations so examiners can award marks throughout the question and potentially award ECF if possible. It is very difficult to do this when students do not show work clearly.

(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

A well answered question.

(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

Only very weak candidates lost this mark.

(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

Many students appeared to have portion of the answer but not the entire concept. They seemed to have studied previous MS by heart and entered an answer that was mostly correct but didn't address the question. It is important that student understand the material and not try to memorize their way through the topics.

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

Question 20N.3.sl.TZ0.10

Select a Test

Syllabus sections

Confirm action