1.

An element with density 11.2 g cm^(-3) forms a fcc lattice with edge length of 4 xx 10^(-8) cm. Calculate the atomic mass of the element. (Given : N_(A) = 6.022 xx 10^(23) mol^(-1))

Answer»

Solution :d = 11.2 `g//cm^(3)`, Z = 4, a = 4 `xx 10^(-8)` cm
`d = (Z xx M)/(N_(A) xx a^(3))`
11.2 `= (4 xx M)/(6.022 xx 10^(23) xx (4 xx 10^(-8))^(3))`
`M = (11.2 xx 6.022 xx 10^(23) xx 4 xx 10^(-8) xx 4 xx 10^(-8) xx 4 xx 10^(-8))/(4)`
`M = 11.2 xx 6.022 xx 16 xx 10^(-1)`
M = 107.9 g `mol^(-1)` or 107.9 u


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