Question

Q.18. An element with molar mass 27 g/mol forms cubic unit cell with edge length of 405 pm. If density of the element is 2.7 g/cm", What is the nature of cubic unit cell? (fcc or ccp)​

Answer 1

Your Answer is FCC or CCP

Solution:

a= 405pm

density =2.7g/cc

N is 6.022×10²³

d= (Z×M) / (N×a³)

you have to find Z

Z = (2.7×10³) (4.5×10^-10)³(6.022×10²³)/0.027

Z= 4

Number of Atom is 4

The cell is FCC or CCP

Hope it will help you....

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