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 × 103 kg m −3 , what is the nature of the cubic unit cell?

Answer 1

Hey there!

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Answer :

Given that,

M = 2.7 × $10^{-2}$ kg $mol^{-1}$

a = 405 pm

d = 2.7 × $10^{3}$ kg $mol^{-3}$

Z = ?

We know that,

Density, d = $\frac{Z × M}{a^{3} × N_{A}}$

⇒ Z = $\frac{d × a^{3} × N_{A}}{M}$

⇒ Z = $\frac{2.7 × 10^{3} kg m^{3} × (405)^{3} × (10^{-12})^{3}m^{3} × 6.023 × 10^{23}}{2.7 × 10^{-2} kg}$

⇒ Z = 4

∴ It is fcc or ccp.

Hence, the nature of cubic unit cell is fcc or ccp.

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