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An element with molar mass 2.7 xx 10^(-2)" kg mol"^(-1) forms a cubic unit cell with edge length 405 pm. If the density is 2.7 xx 10^(3)" kg m"^(-3), what is the nature of the cubice unit cell? |
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Answer» Solution :`"Density, d"=(zxxM)/(a^(3)xxN_(A))""(rho)=(nM)/(a^(3)N_(A))` `therefore""n=(rhoxxa^(3)xxN_(A))/(M)=((2.7xx10^(3)"kg m"^(-3))(4.05xx10^(-10)m)^(3)(6.022xx10^(23)mol^(-1)))/(2.7xx10^(-2)" kg mol"^(-1))` `=3.99=4` Thus, there are 4 atoms of ELEMENTS present per unit cell, hence, the CUBIC unit cell must be FACE - centred or cubic CLOSE - packed (ccp). |
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