Question

calculate the number of atoms of constituent elements
in 53g of Na2CO3

Answer 1

Molar mass of $Na_{2}CO_{3} = (2 \times 23)+(1 \times 12)+(3 \times16)=106$

53 g $Na_{2}CO_{3} = 0.5 mol = 0.5 \times 6.022 \times 10^{23}$

$=3 \times 6.022 \times 10^{23}$ atoms