CBSE Class 12 Chemistry 2022 Term 1 Paper

Question : 32

Total: 55

An element with density 6gcm−3 forms a fcc lattice with edge length of 4×10−8‌cm. The molar mass of the element is (NA=6×1023mol−1)

Solution:

Number of atoms per unit cell of FCC =4
Z=4
We have given -
d=6g∕cm3,
a=‌ edge length ‌=4×10−8‌cm
NA=6×1023
d=‌

a3NA

M=‌ molar mass of element. ‌
∴6g∕cm3=‌

4×M

(4×10−8)3cm3×6×1023mol−1

⇒6×(64×10−24)×6×1023=4M
⇒M=‌

6×6×64×10−1

g∕mol2
∴M=57.6g∕mol2

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