What mass of Ca will contain the same number of atoms as are present in 3.2 g of S?

1 mole of Sulphur weighs 32 g and contains 6.02 x 10²³ atoms

∴ 3.2 g of Sulphur will contain = $\dfrac{6.02 \times 10^{23}}{32} \times 3.2$

= 6.02 x 10²² atoms.

6.02 x 10²³ atoms of Ca weighs = 40 g

∴ 6.02 x 10²² atoms of Ca will weigh = $\dfrac{40}{ 6.02 \times 10^{23}}$ x 6.02 x 10²² = 4 g.