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An element crystallizes in fcc lattice. If the edge length of the unit cell is 408.6 pm and the density is 10.5 g cm^(-3). Calculate the atomic mass of the element. |
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Answer» Solution :`a=408.6 xx 10^(-12)m = 408.6 xx 10^(-10)cm`. `d=10.5 g cm^(-3)` `N_(A)= 6.022 xx 10^(23)"mol"^(-1)` Z = 4 `:.` The CRYSTAL is fcc lattice `M=(DA^(3)N_(A))/(Z)=(10.5 g cm^(-3)xx (408.6xx10^(-10) cm)^(3) xx 6.022 xx 10^(23) "mol"^(-1))/(4)` `=107.8"g mol"^(-1)` * Atomic MASS of the element = 107.8 u. |
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