Chapter U3.64, Problem 3E

(a)

Interpretation Introduction

Interpretation:

The number of atoms of each gas, Ne, Ar and Xe needs to be calculated at STP.

Concept Introduction:

STP is a short form of standard temperature and pressure. The value for standard pressure and temperature is 1 atm and 273.15 K respectively.

At STP 1 mol of any gas occupies 22.4 L.

Answer to Problem 3E

Number of atoms in 22.4 L of each gas is $6.023\times {10}^{23}$ .

Explanation of Solution

At STP, the number of moles of gas having 22.4 L of gas is 1 mol. Since, 1 mol gas must contain Avogadro’s number of molecules and the given gases, Ne, Ar and Xe are monoatomic so the number atom and molecule are same for each gas.

Thus, all the gas must contain $6.023\times {10}^{23}$ number of atoms.

b)

Interpretation Introduction

Interpretation:

The number of density of atoms of each gas, Ne, Ar and Xe needs to be calculated at STP.

Concept Introduction:

Number density of a gas sample is the number of gas molecules present in certain volume of gas sample. It can be also expressed in mol/L.

Answer to Problem 3E

Number density for Ne, Ar and Xe will be same which is 0.045 mol/L.

Explanation of Solution

At STP, the number of moles of gas having 22.4 L of gas is 1 mol.

The number of density of atoms is calculated using formula:

  $\text{number of density of atoms = }\frac{\text{n}}{\text{V}}$

Where n is number of moles and V is volume.

Since, the number of moles of each gas is 1 and volume is 22.4 L so,

density (n/V) $=\frac{1}{22.4}$ mol/L $=0.045$ mol/L.

c)

Interpretation Introduction

Interpretation:

The sample that contains largest mass needs to be determined amongst the following:

Ne, Ar and Xe

Concept Introduction:

The formula used to determine the mass of a sample is:

  $\text{Mass of a sample= number of mol of sample × Molecular weight of sample }$

Answer to Problem 3E

22.4 L Xe will have the largest mass.

Explanation of Solution

At STP, the number of moles of gas having 22.4 L of gas is 1 mol.

Molecular weight of Ne, Ar and Xe are 20 g/mol, 40 g/mol and 131 g/mol respectively.

Thus, mass of 1 mol Ne $\text{=1 mol×20 g/mol =20 g}$ .

Mass of 1 mol Ar $\text{=1 mol×40 g/mol = 40 g}$ .

Mass of 1 mol Xe $\text{=1 mol×131 g/mol = 131 g}$ .

d)

Interpretation Introduction

Interpretation:

The sample that contains largest density needs to be determined amongst the following:

Ne, Ar and Xe

Concept Introduction:

The ratio of mass and volume of a substance is said to be its density. Mathematically,

  $\text{Mass density = }\frac{\text{mass(m)}}{\text{volume(V)}}$

Answer to Problem 3E

Xe has maximum mass density.

Explanation of Solution

At STP, the number of moles of gas having 22.4 L of gas is 1 mol.

Molecular weight of Ne, Ar and Xe are 20 g/mol, 40 g/mol and 131 g/mol respectively.

Thus, mass of 1 mol Ne $\text{=1 mol×20 g/mol =20 g}$ .

Mass density $\text{=}\frac{\text{20}}{\text{22}\text{.4}}\text{ g/L=0}\text{.89 g/L}$

Mass of 1 mol Ar $\text{=1 mol×40 g/mol = 40 g}$ .

Mass density $\text{=}\frac{\text{40}}{\text{22}\text{.4}}\text{ g/L=1}\text{.78 g/L}$ .

Mass of 1 mol Xe $\text{=1 mol×131 g/mol = 131 g}$ .

Mass density $\text{=}\frac{131}{\text{22}\text{.4}}\text{ g/L=5}\text{.85 g/L}$ .