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)
Solution and Explanation
The correct answer is 22.
Given:
M is body centered cubic, so Z = 2.
Let the mass of 1 atom of M be A.
Edge length = 300 pm
Density = \(6 \, \text{g/cm}^3\)
Using the formula for density:
\(6 \, \text{g/cm}^3 = \frac{Z \times A}{(300 \times 10^{-10})^3} = \frac{2 \times A}{27 \times 10^{-24}}\)
Now, solving for A:
\(A = 81 \times 10^{-24} \, \text{g}\)
Atoms of M = \(\frac{\text{Total mass}}{\text{Mass of one atom}}\)
= \(\frac{180}{81 \times 10^{-24}}\)
= \(22.22 \times 10^{23}\)
≅ \(22 \times 10^{23}\)
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Based on the nature of the order that is present in the arrangement of their constituent particles solids can be divided into two types;
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