Question

1. 7.23 × 1024 Al atoms = grams Al

Answer 1

Solution:

$moles= \frac{atoms}{Avagadro \;number}$

Moles of Al $=\frac{7.23 \times10^{23}}{6.023 \times 10^{23}}=12 (\frac{mol}{litres})$

Weight of Al = $moles \times molecular\; weight = 12 \times 27 = 324 grams$

Answer: 324 grams of Aluminium.