Chemistry, asked by vidhi20oct, 1 year ago

chromium crystallises in a bcc lattice whose density is 7.2g/cm³. The length of edge of unit cell is 288.4pm . Calculate the avagadro's number

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✍Formula units per unit cell Z = 2 for BCC cubic unit cell lattice parameter a = 288 pm = 288 x10-8 cm Volume V = a3 =2.39X10-23cm3 Density d = 7.2g/cm3 NA = Avogadro constant = 6.022x10²³ Molecular mass M =? We know that Density d = ZM/NA X a3 M = dxNA x a3/Z On Substituting values M= 7.2g/cm3 x(6.022x10²³)X (6.022x10²³)/2 = 51.8gmol-1 51.8 g of element contains 6.022X1023 208g of this element contains=? = 6.022X1023X208/51.8 =2.42X1024 atoms.

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chromium crystallises in a bcc lattice whose density is 7.2g/cm³. The length of edge of unit cell is 288.4pm . Calculate the avagadro's number ​

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chromium crystallises in a bcc lattice whose density is 7.2g/cm³. The length of edge of unit cell is 288.4pm . Calculate the avagadro's number