Question

the density of lead is 11.35 g/cm3. determine the mass of the rectangular block of lead.
continue from the last question

Answer 1

hey mates your answer is here

Solution :

For a fcc unit cell ,

Density =4×molarmassNA×a3

=4×207gmol−16.02×1023mol−1×a3

=11.35gcm−3

Then

a=(4×2076.02×1023×11.35cm−3)1/3

=4.95×10−8cm

a2–√=4r

From geametry

r=4.95×10−8cm22–√

=1.75×10−8cm

So the radius of lead atom is 1.75×10−8cm or 175pm

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