Respuesta :
The molar mass M of a sample is the ratio between its mass m and the amount of substance n that the sample contains:
[tex]M=\frac{m}{n}[/tex]On the other hand, the density ρ of a sample is the quotient between the mass of the sample and its volume V:
[tex]\rho=\frac{m}{V}[/tex]Finally, Avogadro's Number N_A is the number of particles present in 1 mol of a substance. Then, the number of molecules N present in a sample of n mol is:
[tex]N=n\cdot N_A[/tex]Use these relations to answer the questions.
1)
Isolate n from the first equation and replace the mass of 1.00kg and the molar mass of 175 g/mol to find the amount of substance in 1.00 kg:
[tex]\begin{gathered} n=\frac{m}{M} \\ =\frac{1.00\operatorname{kg}}{175\frac{g}{mol}} \\ =\frac{1000g}{175\frac{g}{mol}} \\ =5.71\text{mol} \end{gathered}[/tex]2)
The volume occupied by a sample of mass m is:
[tex]V=\frac{m}{\rho}[/tex]On the other hand, the mass of one molecule can be found writing it in terms of the amount of substance and the molar mass:
[tex]\begin{gathered} m=M\cdot n \\ \Rightarrow V=\frac{M}{\rho}\cdot n \end{gathered}[/tex]Finally, the amount of substance made up by N molecules is given by:
[tex]n=\frac{N}{N_A}[/tex]Then, the volume occupied by N molecules of a substance with molar mass M and density ρ is:
[tex]V=\frac{M}{\rho}\cdot\frac{N}{N_A}[/tex]Replace M=175 g/mol, ρ=9300 kg/m^3, N=1 and N_A=6.02*10^23 to find the volume of 1 molecule:
[tex]\begin{gathered} V=\frac{175\frac{g}{mol}}{9300\frac{\operatorname{kg}}{m^3}}\cdot\frac{1}{6.02\times10^{23}\cdot\frac{1}{mol}} \\ =\frac{0.175\frac{kg}{mol}}{9300\frac{\operatorname{kg}}{m^3}}\cdot\frac{1}{6.02\times10^{23}\cdot\frac{1}{mol}} \\ =3.126\times10^{-29}m^3 \end{gathered}[/tex]3)
Assuming that all particles are arranged as contiguous cubes with the same volume, then the spacing between moecules is equal to the sidelength of those cubes. Since each cube has a volume of 3.126*10^-29 m^3, then the distance between adjacent molecules would be the cubic root of that amount:
[tex]\begin{gathered} L=\sqrt[3]{V} \\ =\sqrt[3]{3.126\times10^{-29}m^3} \\ =3.15\times10^{-10}m \end{gathered}[/tex]Therefore, there are 5.71 mol in 1.00 kg of the substance, the volume occupied by one molecule of the substance is 3.126*10^-29 m^3 and the spacig between adjacent molecules of the substance is approximately 3.15*10^-10 m.
