DP Chemistry (last assessment 2024)

Question 20N.3.hl.TZ0.b

Date	November 2020	Marks available	[Maximum mark: 2]	Reference code	20N.3.hl.TZ0.b
Level	hl	Paper	3	Time zone	TZ0
Command term	Calculate	Question number	b	Adapted from	N/A

[Maximum mark: 2]

20N.3.hl.TZ0.b

$1.57 %$ 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, ${}^{238}U$ .

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.

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

Options » C: Energy

Question 20N.3.hl.TZ0.b

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