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Coppercrystallisesintoa fccstructureand theunitcellhas lengthof edge 3.61 xx 10^(-8) cm.Calculate thedensity of thecopperAtomicmass ofcopperis 63.5 g mol^(-1)

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Solution :Given : Crystallinestructureof Cuis FCC
Edgelength=a= `3.61 xx 10^(-8) cm`
Atomicmass of Cu=63.5 g `mol^(-1)`
Avogadro number`=6.022 xx10^(23)mol^(-1)`Density = d= ?
In fcc structurethere are8 Cuatomsat 8corners and6Cuatomsat 6facecentres .
`:. ` Totalnumber of Cu ATOMS `=(1)/(8) xx 8 + (1)/(2) xx6 = 1 + 3 =4`
Massof Cu atoms `=4 xx1.054 xx 10^(-22)= 4.216 xx 10^(-22)g`
Mass ofunit cell= Massof 4 Cuatoms`= 4.216 xx 10^(-22) g`
Volumeof unitcell =`a^(3)(3.61 xx 10^(-8))^(3) = 4.7 xx 10^(-23) cm^(3)`
Densityof unit cell`= ("massof unit cell" )/( "volume ofunit cell" ) `
`:.d = (4.216xx 10^(-23))/( 4.7 xx 10^(-23))= 8.97 g cm^(-3)`


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