Question

How many sodium atoms are present in 426g of sodium sulphate.

Answer 1

Mass of 1 mole of sodium sulphate = 142 g
No of sodium sulphate molecules in 426g $=\frac{426}{142}\times 6.022\times {10}^{23}$

No of sodium atom in 1 molecule of sodium sulphate (ie. $N{a}_{2}S{O}_4$) = 2
No of sodium atoms in 426g of sodium sulphate $=2\times \frac{426}{142}\times 6.022\times {10}^{23}$

No of sodium atoms in 426g of sodium sulphate = $3.61\times {10}^{24}$