wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Given that density of NaCl is nearly 2×103 kg m3 and molecular weight is 58.5×103 kg mol1. Then calculate the value of a3 for unit cell of sodium chloride.
Here a is edge length of the fcc unit cell.

Take NA=6×1023

A
1.95×1028 m3
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
3.5×1027 m3
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
0.4×1030 m3
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
4.44×1029 m3
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A 1.95×1028 m3
NaCl is face-centred cubic lattice so that number of NaCl molecules in a unit cell is (Z)=4.
We know,
Density of the unit cell,
ρ=Z×MNA×a3
where,
Z=No. of atoms in a unit cellM=Molar massNA=Avagadro number
a=length of the unit cell

a3=M Zρ NA
Substituting the given values, we get
a3=4×58.5×1032×103×6×1023

a3=1.95×1028m3


flag
Suggest Corrections
thumbs-up
4
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon