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Chemistry Question on The solid state

An element M crystallises in a body centred cubic unit cell with a cell edge of 300 pm. The density of the element is 6.0 g cm–3. The number of atoms present in 180 g of the element is ______× 1023. (Nearest integer)

Answer

The correct answer is 22
M is body centred cubic, ∴ Z=2
Let mass of 1 atom of M is A
Edge length =300pm
Density =6g/cm36 g/cm^3
∴ 6g/cm3=Z×A300×10−103=2×A27×10−246 g/cm^3=\frac{Z×A}{300×10^{-10^3}}=\frac{2×A}{27×10^{-24}}

A=81×10−24gA=81×10^{-24} g
Atoms of M=Total  massMass  of  one  atom\frac{Total \space mass}{Mass\space of \space one \space atom}
= 18081×10−24\frac{180}{81×10^{-24}}
= 22.22×102322.22×10^{23}
≅ 22×102322×10^{23}