DP Chemistry (last assessment 2024)

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Question 18N.3.hl.TZ0.5

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Date November 2018 Marks available [Maximum mark: 6] Reference code 18N.3.hl.TZ0.5
Level hl Paper 3 Time zone TZ0
Command term Calculate, Determine, State Question number 5 Adapted from N/A
5.
[Maximum mark: 6]
18N.3.hl.TZ0.5

A representation of the unit cell of gold is shown.

(a.i)

State the name of the crystal structure of gold.

[1]

Markscheme

face-centred cube/fcc

OR

cubic close packed/ccp ✔

(a.ii)

Calculate the number of atoms per unit cell of gold, showing your working.

[2]

Markscheme

1 2  «atom per face» × 6 «faces per cube» × 3 «atoms» AND 1 8  «atom per corner» × 8 «corners per cube» = 1 «atom» ✔

«atoms per unit cell = 3 + 1 =» 4 ✔

 

Award [1 max] for “4” without working shown.

(b)

The edge length of the gold unit cell is 4.08 × 10‒8 cm.

Determine the density of gold in g cm‒3, using sections 2 and 6 of the data booklet.

[3]

Markscheme

«4 atoms per unit cell»

mass of 4 atoms « = 4 × 196.97 g mo l − 1 6.02 × 10 23 mo l − 1 = » 1.31 × 10–21 «g»

volume of unit cell «= (4.08 × 10‒8)3 cm3» = 6.79 × 10–23 «cm3»

density = « 1.31 × 10 − 21 g 6.79 × 10 − 23 c m 3 » = 1.93 × 101/19.3 «g cm‒3»

 

Award [3] for correct final answer.

 

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