CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

An element with a molar mass of 2.7×10-2 kgmol-1 forms a cubic unit cell with an edge length of 405pm. If its density is 2.7×103 kgm-3, the radius of the element is approximately ______ 10-12m (to the nearest integer).


Open in App
Solution

The radius of the element can be calculated as follows:

Step 1:

Density=ZMNAa3 Z= The number of atoms present in the unit cell.

M =Molar mass

=2.7×10-2kg.mol-1

Na=6.023×1023mol-1

a =edge length of unit cell

=405pm

=405×10-12m

Density =2.7×103kgm-3

Step 2:

So, Density =Z×2.7×10-26.023×1023×405×10-123

Z=6.023×405×405×405×10-8

=4

So we can say, this is a face-centered cubic system.

Step 3:

For face centered cubic system:

4r=a×2

2r×2=405pm2r×2=405×10-12mr=143×10-12m

Hence, the radius of the element is approximately 143 ×10-12m.


flag
Suggest Corrections
thumbs-up
20
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Density
CHEMISTRY
Watch in App
Join BYJU'S Learning Program
CrossIcon