Question

Silver crystallises in fcc lattice. If edge length of the cell is $4.077\times {10}^8$ cm and density is $10.5$ g cm${}^{-3}$, calculate the atomic mass (g) of silver.

• A. $53.54$
• B. $107.09$
• C. $115.54$
• D. $232.23$

Answer 1

Answer: (B)

$107.09$

As we know, density $=\frac{zM}{V{N}_o}$, where n is the number of atoms per unit cell.

Here, $z=4$ for fcc lattice

$10.5=\frac{4\times M}{(4.077\times {10}^{-10}{)}^3(N_o)}$

$⟹M=107$ g

Hence, the correct option is $\text{B}$