Question

.An element with molar mass 2.7×10 -2 kg mol -1 forms a cubic unit cell with edge length 405 pm. If its density is 2.7×10 3 kg m -3 , what is the nature of the cubic unit cell?

Answer 1

Answer:

We know that,

Density=Rank×Molecular mass/Volume×Avagrado's no.

Here, Density =2.7×10^3 kg/m^3

Molecular mass=2.7×10^(-2) kg/mol

Volume={405×10^(-15)}^3

Avagrado's no.=6.022×10^23

2.7×10^3=Rank×2.7×10^(-2)/{405×10^(-15)}^3×6.022×10^23

Rank=2.7×10^3×{405×10^(-15)}^3×6.022×10^23/2.7×10^(-2)

Rank=4

The nature of the cubic unit cell is Face Centred Unit Celll

Answer 2

Answer:

Four atoms of the element are present per unit cell.

Hence, the unit cell is face-centred cubic (fcc) or cubic close-packed (ccp).  

Explanation:

It is given that density of the element, d = 2.7 × 10³ kg $m^{-3}$  

Molar mass, M = 2.7 × $10^{-2}$ kg $mol^{-1}$

Edge length, a = 405 pm = 405 × $10^{-12}$ m    = 4.05 × $10^{-10}$ m

Avogadro’s number, NA = 6.022 × $10^{23}$ $mol^{-1}$

Applying the relation,

$d= \frac{z.M}{a^{3}.N_{A} }$

$z= \frac{d.a^{3}N_{A} }{M}$

z = 2.7 × 10³ kg $m^{-3}$ × (4.05 × $10^{-10}$ m)³ ×6.022 × $10^{23}$ $mol^{-1}$  ÷ 2.7 × $10^{-2}$ kg $mol^{-1}$

z = 4.004

z = 4