Question

The arrangement of ion A and B is shown in cubic figure.
0 →A
•→B
Mass of A atom = 1 amu
Mass of B atom = 2 amu
Edge length = 1 nm
The density of above unit cell (in g/cm°) is (approx.)

Answer 1

Answer:

8.35×10^-3 is the answer

Answer 2

8.3 x $10^{-3}$

Explanation:

Considering the given factors:

A = 1 amu = 1.66 x $10^{-24} g$

B = 2amu = 3.32 x $10^{-24} g$

Z = 1 (for cubic)

a = 1nm = $10^{-7} cm$

M = 1.66 + 3.32

   = 4.98 x $10^{-24}$

formula for density

d= $\frac{ZM}{a^{3}N_{A} }$

Putting up the values

d = $\frac{1* 4.98* 10^{-24} }{10^{-7^{3} } N_{A} }$

   = 8.3 x $10^{-3}$