# Chapter 16 - A Macroscopic Description of Matter - Exercises and Problems: 34

The element's atomic mass number is 56

#### Work Step by Step

Let's assume that we have 1 mole of atoms. We can find the number of atoms along each side of the cube. $(6.02\times 10^{23})^{1/3} = 8.44\times 10^7~atoms$ We can find the length $L$ of each side of the cube. $L = (8.44\times 10^7~atoms)(2.27\times 10^{-10}~m)$ $L = 0.01916~m$ We can find the volume of the cube. $V = L^3$ $V = (0.01916~m)^3$ $V = 7.034\times 10^{-6}~m^3$ We can find the mass of the cube. $m = \rho~V$ $m = (7950~kg/m^3)(7.034\times 10^{-6}~m^3)$ $m = 0.0559~kg$ We can find the mass $m_a$ of each atom. $m_a = \frac{m}{6.02\times 10^{23}}$ $m_a = \frac{0.0559~kg}{6.02\times 10^{23}}$ $m_a = 9.236\times 10^{-26}~kg$ We can find the element's atomic mass number. $A = \frac{m_a}{1.66\times 10^{-27}~kg}$ $A = \frac{9.236\times 10^{-26}~kg}{1.66\times 10^{-27}~kg}$ $A = 56$ The element's atomic mass number is 56.

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