Question

An element crystallizes in a body-centered cubic lattice. The edge of the unit cell is 3.37 Å in length, and the density of the crystal is 7.88 g/cm3 .

Calculate the atomic weight of the element.

Express the atomic weight in grams per mole to three significant digits.

Answer #1

Answer:-

given that it is a body-centered cubic lattice. So "n" = 2.

Edge length of unit cell (a) = 3.37 A^{0}

density of the crystal (d) = 7.88 g//cm^{3}

Now we have to find atomic weight of the element (m) = ?

Density (d) = [ nm / Na^{3} ]

7.88 = [ (2m) / {(6.023
10^{23})
(3.3710^{-8})^{3}}
]

m =( 7.88 6.023
10^{23} 38.2727
10^{-24}) / 2

m =( 1878.8987 10^{-1}
) / 2

m = 187.88987 / 2

m = 93.944935 g/mole

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