Chapter 14, Problem 108A

Interpretation Introduction

Interpretation:

The mole fraction of each gas have to be calculated.

Concept introduction:

Mole fraction: Mole fraction defined as the number of moles of one component divided by total the number of moles in the mixture.

The mole fraction for the solute( ${\chi }_A$ )and the mole fraction for the solvent ( ${\chi }_B$ )can be expressed as follows:Mole fraction ${\chi }_A=\frac{n_A}{ n_A+n_B}$ and ${\chi }_B=\frac{n_B}{ n_A+n_B}$ Where, n_A and n_Brepresent the number of moles of soluteand solvent respectively.

Answer to Problem 108A

The mole fraction of N₂, O₂ and Ar in the sampleare 0.804, 0.189 and 0.0072 respectively.

Explanation of Solution

Data given:

The solution contains N₂ = 78.0%, O₂ = 21.0%and Ar = 1.0%

Let assume, the total mass of the sample = 100 gm

So, the mass of N₂ = 78.0 gm, O₂ = 21.0 gm and Ar = 1.0 gm

$\mathrm{Number}ofmolesof{N}_2=\frac{78gm}{28gm/mol}=2.79mole$

(Note: Molecular weight of N₂ = 28 gm/mol)

$\mathrm{Number}ofmolesof{O}_2=\frac{21gm}{32gm/mol}=0.656mole$

(Note: Molecular weight of O₂ = 32 gm/mol)

$\mathrm{Number}ofmolesofargon=\frac{1.0gm}{39.95gm/mol}=0.025mole$

(Note: Molecular weight of argon = 39.95 gm/mol)

$\begin{array}{l}Molefractionof{N}_2=\frac{Numberofmolesof{N}_2}{Numberofmolesof{N}_2+Numberofmolesof{O}_2+Numberofmolesofargon}\\ =\frac{2.79mol}{(2.79mol+0.656mol+0.025mol)}=0.804\end{array}$

$\begin{array}{l}Molefractionof{O}_2=\frac{Numberofmolesof{O}_2}{Numberofmolesof{N}_2+Numberofmolesof{O}_2+Numberofmolesofargon}\\ =\frac{0.656mol}{(2.79mol+0.656mol+0.025mol)}=0.189\end{array}$

$\begin{array}{l}Molefractionofargon=\frac{Numberofmolesofargon}{Numberofmolesof{N}_2+Numberofmolesof{O}_2+Numberofmolesofargon}\\ =\frac{0.025mol}{(2.79mol+0.656mol+0.025mol)}=0.0072\end{array}$