At a certain temperature and pressure an element has the simple body-centred cubic unit cell, depicted below. The corresponding density is 7.875 g cm-3 and the atomic radius is 1.241 Å. Calculate (and enter) the atomic mass for this element (in amu).

88.95amu

Explanation:

Using D = M/V

But M = Z× AW/A (for a unit cell)

where Z is the number of atoms inside unit cell,

AW is the atomic weight of the atom

A is the Avogadro Const = 6.022 × 10^23 mol^-1

Z = 2: 8 × 1/8 (atoms on corners: only one eighth inside cell) and there is one atom completely inside cell for a total of 2.

First convert 1.780 Å to cm:

1.780 Å = 1.780 x 10-8 cm

Then calculate d, (the edge length of the unit cell ) using Pythagorean Theorem

d2 + (d2)2 = (4r)2 Pythagorean Theorem

3d2 = 16r2

d2 = (16/3) (1.780 x 10-8 cm)2

d = 4.11 x 10-8 cm

 Then Calcuate the volume of the unit cell:

V=a3 where a3 is the side of the unit cell

(4.11 x 10-8 cm)3

= 6.946 x 10-23 cm3

the mass inside the unit cell is given by

(6.946 x 10-23 cm3) × 4.253 g/cm3

= 2.95 x 10-22 g

Use proportionality to calculate the atomic mass:

2.95 x 10-22 g = 2 atoms

 'x'         =      6.022 x 1023 mol-1 (cross multiply and find x)

x = 88.95 amu)

the electrical  conductivity  of asolution  of an electrolyte is measured by determining the resistance of thesolution  between two flat or cylindrical electrodes separated by a fixed distance.

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