Question

an element with molar mass 2.7 ×10^-2 kg mol^-1forms a cubic unit cell with edge length 405 PM. if its density is 2.7×10^3kgm^-3 .what is the radius of the cubic unit cell??​

Answer 1

Answer:

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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