Answer to Question #85713 in Classical Mechanics for Basheer Ahmad

Question #85713
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?
1
Expert's answer
2019-03-05T11:07:33-0500
N/M=(6.021023)/27N/M=(6.02 ∙10^{23})/27

n/m=N/Mn/m=N/M

Thus,


n=mN/M=ρVN/M=(2.7)(0.2)(6.021023)/27=1.21022n=m N/M=ρV N/M=(2.7)(0.2)(6.02 ∙10^{23})/27=1.2∙10^{22}



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