Saved Bookmarks
| 1. |
Silver crystallises in fcc lattice. If edge length of the cell is 4.07 xx 10^(-8) cm and density is 10.5 "g cm"^(-3), calculate the atomic mass of silver. |
|
Answer» Solution :Edge length `(a) = 4.07 xx 10^(-8)` CM `therefore` Volume of unit CELL `= a^3 = (4.07)^3 xx 10^(-24) "cm"^3` Density `(d) = 10.5 "g cm"^3` Atomic Mass (M) = ? NUMBER of ATOMS PER unit cell (Z) = 4 `therefore M= (d xx N_A xx a^3)/(Z)` `therefore M = (10.5 xx 6.022 xx 10^23 xx (4.07)^3 xx 10(-24))/(4)` `therefore` M= 107.12 g/mol . |
|