Q A A-1 1/8 A-2 1 A-3 8r^3 A-4 2r A-5 1.91 A-6 6 Q A B-1 1 B-2 2 B-3 B-4 B-5 2.9
ID: 1059658 • Letter: Q
Question
Q
A
A-1
1/8
A-2
1
A-3
8r^3
A-4
2r
A-5
1.91
A-6
6
Q
A
B-1
1
B-2
2
B-3
B-4
B-5
2.94
B-6
8
Q
A
C-1
1/2
C-2
4
C-3
C-4
C-5
1.91
C-6
6
Q
A
E-1
Both the sphere and hexagonal will be 12
E-2
It will be
E-3
You can’t pack anything better than hexagonal and cubic
Q
A
F-1
¼ Na and 1 cubic, 6 ½ Cl, unit cell 1/8
F-2
Na ¼ (12) +1= 4 Na
Cl ½ 96) + 1/8 (8)= 4 Cl
F-3
2d because it is its double
F-4
Ions of Na=6 , and ions of Cl=6
the first image is the question and the rest it is just information to help you to answer
Q
A
A-1
1/8
A-2
1
A-3
8r^3
A-4
2r
A-5
1.91
A-6
6
Crystal Structure X-ray diffraction studies show that the edge length of the unit cell in crystalline sodium chloride is 5.638 A. Calculate the internuclear Na+-Cl distance. Calculate the density of crystalline sodium chloride. Compare your answer with the experimental value obtained from the Handbook of Chemistry and Physics.Explanation / Answer
NaCl crystallizes as face centered cubic (fcc) structure .
Internuclear distance for a fcc crystal d = a / Square root of 2
a = 5.6358 Å (given), square root of 2 = 1.4142
d = 5.6358 Å / 1.4142 = 3.9852 Å.
For determining the density of a unit cell of NaCl we need to apply the formula, Density = Mass / Volume.
Volume of the unit cell will be = a3 = (5.6358 Å)3 = (5.6358 x 10-8 cm)3 = 1.7900 x 10-22 cm3
There are 4 NaCl molecule present in one unit cell (as it crystallises in face centered cube)
Molar mass of NaCl = 58.44 g/mole
Mass of one molecule of NaCl = (58.44 g/ mol) / (6.022 x 1023 molecules / mol) = 9.7 x 10-23 g/molecule
Mass of 4 molecules of NaCl = 9.7 x 10-23 g/molecule * 4 molecules = 3.88 x 10-22 g
Density of unit cell of NaCl = Mass / Volume = 3.88 x 10-22 g / 1.7900 x 10-22 cm3 = 2.167 g/cm3.
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.