An element with density `11.2 g cm^(-3)` forms a f.c.c. lattice with the edge length of `4xx10^(-8)` cm. Calculate the atomic ma

Correct Answer - `107 "g mol"^(-1)`
`rho=(ZxxM)/(a^3xxN_A)` For element with f.c.c. lattice , Z=4
`therefore M=(rhoxxa^3xxN_A)/Z=((11.2 g cm^(-3))(4xx10^(-8) cm)^3xx(6.022xx10^23 mol^(-1)))/4=107.9 "g mol"^(-1)`

An element with density `11.2 g cm^(-3)` forms a f.c.c. lattice with the edge length of `4xx10^(-8)` cm. Calculate the atomic mass of the element. (Gi

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