Q12 of 26 Page 1

An element ‘X’ (At. mass = 40 g mol–1) having f.c.c. structure, has unit cell edge length of 400 pm. Calculate the density of ‘X’ and the number of unit cells in 4 g of ‘X’. (NA = 6.022 × 1023 mol–1 )

Given:

Atomic mass of element(m)=40 gmol-1 length of unit cell(a)=400 pm=4× 10-8 cm; Z=4(Fcc).


Now density is given by the formula



Where Z is the number of an atom in the unit cell


M is the molecular mass


NA is the Avogadro number


V is the volume of the unit cell.


V=a3=(4× 10-8)3 cm =64× 10-24 cm3


Putting the values



Density, D=160/38.5=4.1 g cm-3


More from this chapter

All 26 →
10

Complete and balance the following chemical equations :

(a) Fe 2+ + MnO4- + H+ → �


(b) MnO4-+ H 2O + I – →

 11

Give reasons for the following :

(a) Measurement of osmotic pressure method is preferred for the determination of molar masses of macromolecules such as proteins and polymers.


(b) Aquatic animals are more comfortable in cold water than in warm water.


(c) Elevation of the boiling point of 1 M KCl solution is nearly double than that of 1 M sugar solution.

 13

A first order reaction is 50% completed in 40 minutes at 300 K and in 20 minutes at 320 K. Calculate the activation energy of the reaction.

(Given: log 2 = 0.3010, log 4 = 0.6021, R = 8.314 JK–1 mol–1)

 14

What happens when

(a) a freshly prepared precipitate of Fe(OH)3 is shaken with a small amount of FeCl3 solution?


(b) persistent dialysis of a colloidal solution is carried out?


(c) an emulsion is centrifuged ?