Question

Calculate the number of atoms in $39.4g$ gold. Molar mass of gold is $197g/mole$.

Answer 1

Given,

Mass of the gold, $m=39.4g$

Molar mass of the gold, $M=197g/mole$

Avogadro's number $N_A=6.023\times {10}^{23}$ atoms.

$1$ mole = Avogadro's number of atoms.

Number of molecules $=N_A\times n$

$197g$ of gold contains $6.023\times {10}^{23}$ atoms.

$1g$ of gold contains $\frac{6.023\times {10}^{23}}{197}$ atoms.

$\therefore 39.4g$ of gold contains $\frac{6.023\times {10}^{23}\times 39.4}{197}$
atoms $=1.20\times {10}^{23}$ atoms.

$\text{Final Answer:}$ $1.20\times {10}^{23}$ atoms.