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The density of silver having atomic mass 107.8 gram `"mol"^(-1)` is 10.8 gram `cm^(-3)`. If the edge length of cubic unit cell is `4.05xx10^(-8)cm`, find the number of silver atoms in the unit cell. `(N_(A)=6.022xx10^(23),1A^(@)=10^(-8)cm)` |
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Answer» Given, Atomic mass of `Ag=107.8 g "mol"^(-1)` Density `(rho)=10.8g "cm"^(-3)` Edge length `(a)=4.05xx10^(-8)cm` No. of silver atom `=?` Formula, `rho=(Z xx M)/(a^(3)xx N_(A)),Z=` No. of atoms is unit cell `:.Z=(rhoxx a^(3)xx N_(A))/(M)` `=(10.8xx(4.05xx10^(-8))^(3)xx6.022xx10^(23))/(107.8)` `=(10.8xx66.43xx10^(-24)xx6.022xx10^(23))/(107.8)` `=(4320.45xx10^(-1))/(107.8)` `=(432.045)/(107.8)` `=4.007~=4` 4 atoms of silver in unit cell. |
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