Calculate the number of atoms in 0.9g of aluminium.

The molar mass of aluminium (Al) is given as 27 g/mol.

$$\text{Moles of Al} = \frac{\text{Mass of Al}}{\text{Molar mass of Al}}$$ $$\text{Moles of Al} = \frac{0.9 \text{ g}}{27 \text{ g/mol}}$$ $$\text{Moles of Al} = 0.0333... \text{ mol}$$ $$\text{Moles of Al} = \frac{1}{30} \text{ mol}$$

We are given that 1 mole of aluminium contains $6 \times 10^{23}$ atoms.

$$\text{Number of atoms} = \text{Moles of Al} \times \text{Avogadro's number}$$ $$\text{Number of atoms} = \frac{1}{30} \text{ mol} \times 6 \times 10^{23} \text{ atoms/mol}$$ $$\text{Number of atoms} = \frac{6}{30} \times 10^{23} \text{ atoms}$$ $$\text{Number of atoms} = \frac{1}{5} \times 10^{23} \text{ atoms}$$ $$\text{Number of atoms} = 0.2 \times 10^{23} \text{ atoms}$$ $$\text{Number of atoms} = 2.0 \times 10^{22} \text{ atoms}$$

Comparing this result with the given options:
A. $1.0 \times 10^{21}$
B. $6.6 \times 10^{21}$
C. $2.0 \times 10^{22}$
D. $6.0 \times 10^{22}$

The calculated number of atoms matches option C.

The number of atoms contained in 0.9g of aluminium is $\boxed{\text{C. } 2.0 \times 10^{22}}$.