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Question
An element has a bcc structure with a unit cell edge length of 288 pm. How many unit cells and a number of atoms are present in 200 g of the element? (1.16 × 1024, 2.32 × 1024)
Solution
V = a3 = (288 × 10−10 cm)3 = 2.39 × 10−23 cm3
From the relationship:
Density `= (Z xx M)/(N_Axxa^3)`
Where:
-
Z = 2 (atoms per unit cell for BCC)
-
M = molar mass in g/mol
-
NA = 6.022 × 1023 mol⁻¹
-
a3 = volume of unit cell in cm3
Moles `= 200/M`
Number of atoms = `200/M xx N_A`
`200/M xx 6.022 xx 10^23 = 2.32xx10^24`
`M = (200xx6.022xx10^23)/(2.32xx10^24) = 51.9` g/mol
Number of unit cells = `(2.32 xx 10^24)/2`
1.16 × 1024
Number of unit cells = 1.16 × 1024
Number of atoms = 2.32 × 1024
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