Question

The number of atoms and g-atoms present in:
i. 60 g carbon is $\times {10}^{23}$ and respectively.
ii. 224.4 g $Cu$ is $\times {10}^{23}$ and respectively.

(Atomic Weight of $C$ and $Cu$ are 12 u and 63.6 u, respectively. Avogadro’s number = $6.02\times {10}^{23})$

Answer 1

$\text{g-atom}=\frac{\text{weight}}{\text{gm-atomic weight}}$
$\text{Number of atoms}=\frac{\text{weight}\times N_A}{\text{gm-atomic weight}}$
(a) For 60 g of $C$: g atom =$\frac{60}{12}=5$
Number of atoms: $\frac{60\times 6.02\times {10}^{23}}{12}=30.1\times {10}^{23}$

(b) For 224.4 g of $Cu$: g atom =$\frac{224.4}{63.6}=3.53$
Number of atoms $=\frac{224.4\times 6.02\times {10}^{23}}{63.6}=21.24\times {10}^{23}$