Question

Determine the number of atoms in 1·790 × 10^-6 g of silver (Ag = 108).

Answer 1

Weight of one silver atom $=\frac{\text { at. wt. of silver }}{6 \cdot 02 \times 10^{23}}$

$=\frac{108}{6 \cdot 02 \times 10^{23}} g =17 \cdot 9 \times 10^{-23} g .$

$\because \quad 17 \cdot 9 \times 10^{-23} g$ is the weight of one silver atom

$\therefore$ Number of atoms in $1.790 \times 10^{-6} g =\frac{1.790 \times 10^{-6}}{17 \cdot 9 \times 10^{-23}}$ atoms $=10^{16}$ atoms.