Silver crystallises in the fcc lattice. If edge length of the cell is $4.07 \times 10^{-8} \mathrm{~cm}$ and density is $10.5 \mathrm{~g} \mathrm{~cm}^{-3}$, calculate the atomic mass of silver. - Noon Academy

Silver crystallises in the fcc lattice. If edge length of the cell is $4.07 \times 10^{-8} \mathrm{~cm}$ and density is $10.5 \mathrm{~g} \mathrm{~cm}^{-3}$ , calculate the atomic mass of silver.

Previous year question papers Maths

Silver crystallises in the fcc lattice. If edge length of the cell is $4.07 \times 10^{-8} \mathrm{~cm}$ and density is $10.5 \mathrm{~g} \mathrm{~cm}^{-3}$ , calculate the atomic mass of silver.

Ans: It is given that the edge length, $\mathrm{a}=4.07 \times 10^{-8} \mathrm{~cm}$ and density is $\mathrm{d}=$ $10.5 \mathrm{~g} \mathrm{~cm}^{3}$
As the lattice is foc type, the number of atoms per unit cell, $z=4$ We also know that, $\mathrm{N}_{\mathrm{A}}=6.022 \times 10^{23} \mathrm{~mol}^{-1}$
Using the relation:

$\begin{aligned} d &=\frac{z \mathrm{M}}{\mathrm{a}^{3} \mathrm{~N}_{A}} \\ & \Rightarrow \mathrm{M}=\frac{\mathrm{da}^{3} \mathrm{~N}_{\mathrm{A}}}{\mathrm{Z}} \\ &=\frac{10.5 \mathrm{~g} \mathrm{~cm}^{-3} \times\left(4.077 \times 10^{-8} \mathrm{~cm}\right)^{3} \times 6.022 \times 10^{28} \mathrm{~mol}^{-1}}{4} \\ &=107.13 \mathrm{gmol}^{-1} \end{aligned}$

Therefore, atomic mass of silver $=107.13 \mathrm{u}$

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