Saved Bookmarks
| 1. |
Gold occurs as face centred cube and has a density of 19.30kg dm^(-3). Calculate atomic radius of gold. (Molar mass of Au=197) |
|
Answer» Molar MASS=`197g"mol"^(-1)` Avogadro constant `=N_(A)=6.022xx10^(23)"mol"^(-1)` Atomic radius of Au=? If fcc unit cell, there are 8 ATOMS of Au at 8 corners and 6 atoms at 6 face centres. Number of Au atoms in the unit cell `=(1)/(8)xx8+(1)/(2)xx6` 4atoms Mass of 1Au atom `(197)/(6.022xx10^(23))=3.271xx10^(-22)g` `therefore "Mass of 4 Au atoms"=4xx3.271xx10^(-22)g` `therefore "Mass of unit cell"=1.308xx10^(-22)g` `""=1.308xx10^(-24)kg` Densidy of the unit cell `=("Mass of unit cell")/("Volume of unit cell")` `therefore d=(1.308xx10^(-21))/(a^(3))` `therefore (a^(3)=1.308xx10^(-21))/(d)=(1.308xx10^(-24))/19.3` `=6.77xx10^(-26)dm^(3)` `=6.77xx10^(-23)cm^(3)` `therefore a=(6.77xx10^(-23))^(1//3)=(67.77xx10^(-24))^(1//3)` `""=4.077xx10^(-8)cm` If r is the radius of Au atom, then for fcc unit cell, `r=(a)/(2sqrt2)` `=(4.077xx10^(-8))/(2sqrt2)=1.442xx10^(-8)cm=144.2cm` |
|