Chapter 14, Problem 84A

What is the mole fraction of H 2 S O 4 in a solution containingthe percentage of sulfuric acid and water shownin Figure 14.25?

Interpretation Introduction

Interpretation:

The mole fraction of H₂SO₄ in the sulfuric acid solution is 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 84A

The mole fraction of $H_{2}S{O}_4$ in the sulfuric acid solution is 0.0644.

Explanation of Solution

Given information:

The solution contains:

$H_{2}S{O}_4=27.3\%$ and

$Water=72.7\%$

Let assume, the mass of the solution = 100 gm

So, the mass of H₂SO₄ = 27.3 gm and the mass of water = 72.7 gm

$\mathrm{Number}ofmolesofsulfuricacid=\frac{27.3gm}{98.1gm/mol}=0.278mole$

(Note: Molecular weight of sulfuric acid = 98.1 gm/mol)

$\mathrm{Number}ofmolesofwater=\frac{72.7gm}{18gm/mol}=4.04mole$

(Note: Molecular weight of water = 18 gm/mol)

$\begin{array}{l}Molefractionof{H}_{2}S{O}_4=\frac{Numberofmolesof{H}_{2}S{O}_4}{Numberofmolesof{H}_{2}S{O}_4+Numberofmolesofwater}\\ =\frac{0.278mol}{0.278mol+4.04mol}=0.0644\end{array}$