# Chapter 13 - Problems - Page 498: 40

(a) $2.69\times 10^{25}~molecules/m^3$ (b) The average distance between molecules is $3.34\times 10^{-9}~m$ (c) The total mass is $0.028~kg$ The mass density is $1.25~kg/m^3$

#### Work Step by Step

(a) We can find the number density: $\frac{6.022\times 10^{23}}{0.0224~m^3} = 2.69\times 10^{25}~molecules/m^3$ (b) To find the average distance between molecules, we can assume the molecules are spread evenly in a cube with a volume of $0.0224~m^3$ We can find the length of each side of the cube: $L = (0.0224~m^3)^{1/3} = 0.282~m$ We can find the number of molecules along one side of the cube: $(6.022\times 10^{23})^{1/3} = 8.445\times 10^7~molecules$ We can find the average distance between molecules: $\frac{0.282~m}{8.445\times 10^7} = 3.34\times 10^{-9}~m$ The average distance between molecules is $3.34\times 10^{-9}~m$ (c) A nitrogen atom consists of 7 protons and 7 neutrons. The mass of a nitrogen molecule $N_2$ is $28~u$ Since the mass of one nitrogen molecule is $28~u$, we know the mass of one mole of nitrogen molecules is $28~grams$ which is $0.028~kg$ We can find the mass density: $\frac{0.028~kg}{0.0224~m^3} = 1.25~kg/m^3$

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