Question

A solid cube of aluminum (density 2.7 g/cm3) has a volume of 0.2 cm3. It is known that 27 g of aluminum contains 6.02 x 1023 atoms. How many aluminum atoms are contained in the cube?

Answer 1

$$N/M=(6.02 ∙10^{23})/27$$

$$n/m=N/M$$

Thus,

$$n=m N/M=ρV N/M=(2.7)(0.2)(6.02 ∙10^{23})/27=1.2∙10^{22}$$