The Solid State

Question 241
CBSEENCH12005589

Aluminium forms fcc cubic crystals. The density of aluminium is 2.7 g cm–3. Calculate the length of the edge of the unit cell of Al.

Solution

we have give that aluminium forms Fcc cubic crystal.
density = 2.7g cm-3
Mass of aluminium = 27g
number of unit cell =4 (FCC)
we have find edge length
Thus using formula

d =Z xMa3 x NAputting the value in this equation 2.7 =4 x276.023 x 1023 x a3 a3 = 4 x 27 6.023 x 10 23 x 2.7a =4.05 x 10-8


Question 242
CBSEENCH12005590

The density of KBr is 2.75 g cm–3. The length of the edge of the unit cell is 654 pm. Show that KBr has a fcc structure.

Solution

we know that

d= ZMa3NA

as given that,
density= 2.75 g cm–3
a = 654 pm or a3 = 2.79 x 10-23
mass of KBr is 119 gram



Z =2.75 x6.54 x10-23 x6.022 x1023119 =4

Question 243
CBSEENCH12005591

Tungsten has a density of 19.35 g cm–3 and the length of the side of the unit cell is 316 pm. The unit cell is a body centred unit cell. How many atoms does 50 grams of the element contain ?

Solution

we have given,

density =19.35 g cm-3
edge length =316 pm
Number of unit cell =2 (BCC)
Mass =?

Thus,
Length of edge of the unit cell = 316 pm =316 x 10-10cmvolume of the unit cell = (316 x 10-10cm)3                              = 3.2 x 10-23 cm3/unit cellDensity  =atomic mass x Zunit cell volume  x NAAtomic mass = Density xUnit volume  xNAZ=19.35 x 3.2 x 10-23 x 6.023 x 10 232 =186.5 g/ molAs 186.5 g the element contain NA atoms   = 6.023 x 10 23 atomsso 50g of the element contain NA atoms =6.023 x 1023 x 50186.5=1.614 x 1023 atoms

Question 244
CBSEENCH12005592

In the cubic crystal of CsCl (d = 3.97 g cm–3), the eight corners are occupied by CIwith a Cs+ at the centre and vice versa. Calculate the distance between the neighbouring Cs+ and CI ions. What is the radius ratio of the two ions ? (At. mass of Cs = 132.91 and CI = 35.45).

Solution

In a unit cell there are one Cs and 1 x 8/8 =1 chlorine  (Cl-) such that one CsCl molecule 

therefore 
As we have given 
density = 3.97 g cm-3
Mass of CsCl = 168.36g
Number of unit cell(Z) = 1

apply formula d= Z xMa3 x NA3.97 = 1 x 168.36a3 x 6.023 x 1023a = 4.13 x 10-8 cma =4.13 A0

for a cube of side  length 4.13A0 diagonal

                                    =3a i.e.,3 x 4.13 = 7.15A0

as it is a BCC with  Cs+ at centre radius r+ and Cl- at corner radius r- so, 

2r+ +2r- =7.15 or r+ +r- =3.57A0

such that distance between neighbouring Cs+ and Cl- =3.57A0

now assume two Cl- ion touch with each other so length of unit cell = 2r- =4.13
r- =2.06A0
r+ =3.57 -2.06 =1.51

r+/r-=1.51/2.06 =0.73

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