Discussion Forum

 An element has a bcc structure with cell edge of 288 pm . The density of element is 7.2 g cm^-3. what is the atomic mass of an element ?

Answer:

Explanation:

 Given ,

 Cell in bcc, so Z=2

 Edge length (a) =288 pm

                        =288 X 10^-8 cm

 Density of metal (d) = 7.2 g cm^-3

                  N_A= 6.022 x 10²³

 We know that , density (d)= $\frac{Z \times M}{a^{3}.N_{A}}$

$\therefore M=\frac{d.a^{3}.N_{A} }{Z}$

$=\frac{7.2\times (2.88\times 10^{-8})\times6.022\times 10^{23}}{2}$

$=\frac{7.2\times 23.8878\times 10^{-24}\times6.022\times 10^{23}}{2}$

= 51.78 g mol^-1