Chapter U3.64, Problem 4E

(a)

Interpretation Introduction

Interpretation:

The number of molecules of each gas, H₂, N₂ and CO₂ 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 4E

Number of molecules 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, H₂, N₂ and CO₂ are of 1 mole.

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

b)

Interpretation Introduction

Interpretation :

The number of density of atoms of each gas, H₂, N₂ and CO₂ 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 4E

Number density for H₂, N₂ and CO₂ 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:

H₂, N₂ and CO₂

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 4E

22.4 L CO₂ 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 H₂, N₂ and CO₂ are 2 g/mol, 28 g/mol and 144 g/mol respectively.

Thus, mass of 1 mol H₂$\text{=1 mol×2 g/mol=2 g}$ .

Mass of 1 mol N₂$\text{=1 mol×28 g/mol=28 g}$ .

Mass of 1 mol CO₂$\text{=1 mol×44 g/mol=44 g}$ .

d)

Interpretation Introduction

Interpretation :

The sample that contains largest number of atoms needs to be determined amongst the following:

H₂, N₂ and CO₂

Concept Introduction :

Number of molecules per mol of substance is equal to Avogadro’s number which is equal to $6.023\times {10}^{23}$ .

Atomicity is the number of atoms per molecules of an element.

Number of atoms $\text{=atomicity×number of molecules}$

Answer to Problem 4E

CO₂ has the largest number of atoms.

Explanation of Solution

At STP 22.4 L H₂, N₂ and CO₂ gas is equal to 1 mol.

Number of molecules in each gas sample is $6.023\times {10}^{23}$ .

H₂, N₂ and CO₂have 2,2 and 3 atoms per molecule respectively.

Number of atoms of H₂$\text{=2}\times \text{6}\text{.023}\times {\text{10}}^{\text{23}}\text{.}$

Number of atoms of N₂$\text{=2}\times \text{6}\text{.023}\times {\text{10}}^{\text{23}}\text{.}$

Number of atoms of CO₂$\text{=3}\times \text{6}\text{.023}\times {\text{10}}^{\text{23}}\text{.}$

e)

Interpretation Introduction

Interpretation:

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

H₂, N₂ and CO₂

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 4E

CO₂ 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 H₂, N₂ and CO₂ are 2 g/mol, 28 g/mol and 144 g/mol respectively.

Thus, mass of 1 mol H₂$\text{=1 mol×2 g/mol=2 g}$ .

Mass density $\text{=}\frac{\text{2}}{\text{22}\text{.4}}\text{ g/L=0}\text{.089 g/L}$

Mass of 1 mol N₂$\text{=1 mol×28 g/mol=28 g}$ .

Mass density $\text{=}\frac{\text{28}}{\text{22}\text{.4}}\text{ g/L=1}\text{.25 g/L}$

Mass of 1 mol CO₂$\text{=1 mol×44 g/mol=44 g}$

Mass density $\text{=}\frac{44}{\text{22}\text{.4}}\text{ g/L=1}\text{.96 g/L}$ .