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An elementcrystallises inthe fcccrystallatticeand hasa densityof 10 g cm^(-3)withunitcell edgelengthof 100 pm . Calculatenumberof atomspresentin 1 gof crystal. |
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Answer» Edgelength =a= 100 pm `=1xx 10^(-8)cm` Avogadronumber `= N_(A) = 6.022 xx 10^(23) mol^(-1)` Numberof atom = s PRESENTIN 1 gcrystal = ? LETN be theatomicmass of theelement . Mass of one atom `= (M )/(6.022 xx 10^(23))` Fcc typeunitcellcontains 4 atoms. `:. ` Massof unitcell = Massof 4 atoms `= 4 xx (M)/(6.022 xx 10^(23))` Volumeof unitcell `=a^(3) = (1 xx 10^(-8))^(3) =1 xx 10^(-24) cm^(3)` Densityof the crystal =d= `("Mass ofunit CELL" )/("volumeof unit cell" )` `10= 4 xx M //6.022 xx (10^(23))/(1xx 10^(-24))` `= (4 xx M)/(6.022 xx 10^(23) xx 1 xx 10^(-24)) = (4 M)/(0.6022)` `:. M = (10 xx0.6022)/(4) =1.5 g mol^(-4)` Now1 gramatom= 1mole ofelement=1.5g `:.` 1.5 gelementcontains6.022 `xx 10^(23)` atoms `:.` 1 g element willcontains.` (6.022 xx 10^(23))/(1.5) =4 xx 10^(23) ` atoms |
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