Chapter 13, Problem 13.74P

Interpretation Introduction

Interpretation:

The molality, molarity and mole fraction of ${\text{NH}}_3$ is to be determined.

Concept introduction:

Molality is the measure of the concentration of solute in the solution. It is the amount of solute that is dissolved in one kilogram of the solvent. It is represented by $\text{m}$ and its unit is moles per kilograms. The solute is the substance that is present in a smaller amount and solvent is the substance that is present in a larger amount.

The formula to calculate the molality of the solution is as follows:

$\text{Molality}=\frac{\text{amount}\left(\text{mol}\right)\text{of}\text{solute}}{\text{mass}\left(\text{kg}\right)\text{of}\text{}\text{solvent}}$ (1)

Molarity is defined as the number of moles of solute that are dissolved in one litre of solution. It is represented by $\text{M}$ and its unit is $\text{mol/L}$.

The formula to calculate the molarity of the solution is as follows:

$\text{Molarity}=\frac{\text{amount}\left(\text{mol}\right)\text{of}\text{solute}}{\text{volume }\left(\text{L}\right)\text{of}\text{}\text{solution}}$ (2)

The mole fraction is defined as the ratio of the number of moles of solute to the total number of moles in the mixture. It is represented by $X$.

The formula to calculate the mole fraction is as follows:

$\text{Mole}\text{fraction}\left(X\right)=\frac{\text{amount}\left(\text{mol}\right)\text{of}\text{solute}}{\text{amount}\left(\text{mol}\right)\text{of}\text{solute}+\text{amount}\left(\text{mol}\right)\text{of}\text{solvent}}$ (3)