the density of lead is 11.35 g/cm3. determine the mass of the rectangular block of lead.
continue from the last question
Answers
Answered by
3
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
may be it's helpful for you
please mark me as brainliest ✌️✌️
Ritiksuglan:
please mark me as brainliest ✌️✌️
Similar questions