Chapter 14.2, Problem 29PP

Interpretation Introduction

Interpretation:

Mole fraction of NaOHin aqueoussolution has to be calculated.

Concept introduction:

Mole fraction: It’s another way of expressing concentration of a solution. Mole fraction defined as the number of moles of one component divided by total the number of moles in the mixture. The number of moles of one species in a mixture is always less than the total number of , oles thus, mole fraction has value between 0 to 1. Mole fraction is denoted as ${\chi }_i$ where subscript $i$ represents the particular component. The sum of the mole fractions of the components in a mixture (or a solution) is unity.

The mole fraction for the solvent $({\chi }_A)$ and the mole fraction for the solute $({\chi }_B)$ 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, ? _A and ?_B represent the mole fractions of each substance. n_A and n_Brepresent the number of moles of each substance.

$({\chi }_A+{\chi }_B)=1$

Answer to Problem 29PP

The mole fraction of NaOH is 0.118.

Explanation:

Write the expression of mole fraction:

Mole fraction ${\chi }_A=\frac{n_A}{n_A+n_B}$

Aqueous solution contains 22.8 % NaOH by mass means 100.0 g aqueous solution of NaOH contains 22.8 g NaOH. So, mass of NaOH = 22.8 gm

Mass of water = Mass of solution − Mass of NaOH = (100 − 22.8) gm = 77.2 gm

Number of moles of NaOH (n _NaOH) $=\frac{22.8}{40}$ = 0.57 mol

Calculate the moles of water:

$\begin{array}{c}\text{Number of moles of water }\left({\text{n}}_{\text{water}}\right)\text{ =}\frac{\text{77}\text{.2}}{\text{18}}\\ \text{= 4}\text{.28 mol}\end{array}$

Write the expression for Mole fraction of NaOH

${\chi }_{NaOH}=\frac{n_{NaOH}}{(n_{NaOH}+n_{water})}$

$\begin{array}{l}=\frac{0.57}{(0.57+4.28)}\\ ==0.118\end{array}$

The mole fraction of NaOH is 0.118.

Explanation of Solution

Write the expression of mole fraction:

Mole fraction ${\chi }_A=\frac{n_A}{n_A+n_B}$

Aqueous solution contains 22.8 % NaOH by mass means 100.0 g aqueous solution of NaOH contains 22.8 g NaOH. So, mass of NaOH = 22.8 gm

Mass of water = Mass of solution − Mass of NaOH = (100 − 22.8) gm = 77.2 gm

Number of moles of NaOH (n _NaOH) $=\frac{22.8}{40}$ = 0.57 mol

Calculate the moles of water:

$\begin{array}{c}\text{Number of moles of water }\left({\text{n}}_{\text{water}}\right)\text{ =}\frac{\text{77}\text{.2}}{\text{18}}\\ \text{= 4}\text{.28 mol}\end{array}$

Write the expression for Mole fraction of NaOH

${\chi }_{NaOH}=\frac{n_{NaOH}}{(n_{NaOH}+n_{water})}$

$\begin{array}{l}=\frac{0.57}{(0.57+4.28)}\\ ==0.118\end{array}$

The mole fraction of NaOH is 0.118.