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A body centred cubic element of density, 10.3 g cm^(-3) has a cell edge of 314 pm. Calculate the atomic mass of the element. (N_A = 6.023 xx 10^(23) mol^(-1))

Answer»


Solution :Z=2 for BEC structure , d = 10.3 G `cm^(-3)`, a = 314 pm , `N_A =6.023 xx 10^(23) mol^(-1)`
`M = (N_A xx d xx a^3)/(Z)=(6.023 xx 10^(23) xx 10.3 xx (314)^3 xx 10^(-30))/(2) = 96 g mol^(-1)`


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