(a) Determine the type of cubic lattice to which a given crystal belongs if it has edge length of 290 pm and density is 7.80 g cm–3. (Molecular mass.= 56 g mol–1)

(b) Why does zinc oxide exhibit enhanced electrical conductivity on heating?


we know that

d =ZMa3NA

as we have given 
d= 7.80 g cm–3.
M=56 g mol–1
a= 290 pm or a3 =2.43 x10-23
putting all value in above equation we get,

Z =a3 x d xNAMZ= 2.43 x10-23 x 7.80 x6.022 x102356 =2.03

hence it belong to bcc crystal lattice.

b) Zinc oxide is white in colour at room temperature. On heating it loses electron and turns yellow in colour. 


ZnO Zn2+ +12O2 +2e- 

The electron liberate after heating, can act as charge carrier and thus on heating zinc oxide it exhibit electrical conductivity.

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The compound CuCl has the ZnS (Cubic) structure. Its density is 3.04 g cm–3. What is the length of the edge of unit cell? (At. mass of Cu = 63.5, CI = 35.5) 

Formula mass of CuCl =63.5 +35.5

                                      =  99.0

The number of formula units per cell of ZnS is 4. It has face centred cubic structure.

 we have Z  =4density (d) =3.4 g Cm-3N0 (Avogadro's number) =6.023 x1023Mass =99.0 gramTherefore using formula density =Mass x Zunit cell volume(Cm)3 xNavod.no.d =M x Za3 x N0a3 =99 x43.4 x6.023 x1023a3 =1.932 x10-22cm3unit cell length a= (1.932 x10-22)13takinig log both sidelog a =13log  1.932 x10-22       =13(0.2860-22) =13(-21.7140)       = -7.238 =0.762 x10-8taking antilog a =antilog 0.762 x10-8 =5.758 x10-8 cm = 5.758 x 10-8cm




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An element crystallises in fcc structure with an edge of 200 pm. Calculate density if 200 g of this element contains 24 x 1023 atoms. 

we have 
edge length =200Pm
volume of the unit = ( 200 x10-10 cm)3
                         = 8 x10-24 cm-3

In a FCC unit cell there are four atoms per unit cell.therefore  mass of unit cell  =200 x 424 x1023 =33.3 x10-23gDensity  =mass of unit cell volume of unit cell =33.3 x10-23g8 x10-24cm3  =41.6 g cm-3

density = 41.6 g cm-3.





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Potassium crystallises in a simple cubic unit cell. It has an atomic mass of 209 and its density is 91.5 g m–3. What is the edge length of its unit cell? 

we have given,
Mass= 209 g
Number of  atom per unit cell =1 (Simple cubic)
density =91.5 g m-3
NA =6.023 x1023
edge length of unit cell =?
 
By applying formula


density = Mass x Number of atom per unit cellvolume of unit cell  x avogadro's numberd= M x Za3 x NA91.5 = 209 x 1a3 x 6.023 x 1023a3 =209 x 191.5 x 6.023 x 1023

a= 15.59 x10-8 cm

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Calculate the efficiency (percentage of volume occupied and unoccupied) of packing in case of a metal crystal for simple cubic.


Let ‘a’ be the edge length of the unit cell and r be the radius of sphere.

As sphere are touching each other

Therefore a = 2r

No. of spheres per unit cell = 1/8 × 8 = 1 

Volume of the sphere = 4/3 πr3 

Volume of the cube = a3= (2r)3 = 8r3

∴ Fraction of the space occupied = 1/3πr3 / 8r3 = 0.524

∴ % occupied = 52.4 %



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