Chapter 13, Problem 13.15QP

Interpretation Introduction

Interpretation: A solution has prepared at particular concentration at $\text{20°C}$ which is heated about $\text{70°C}$. What will change along molarity, molality, percent by mass and mole fraction has to be explained.

Concept introduction

Molality (m): Molality is the number of moles of solute present in one kilogram of solvent.

$\text{Molality (m) =}\frac{\text{Number}\text{of}\text{moles}\text{of}\text{solute}}{\text{1kg}\text{of}\text{solvent}}$

Molarity (M): Molarity is number of moles of the solute present in the one liter of the solution.

$\text{Molarity (M) =}\frac{\text{Number}\text{of}\text{moles}\text{of}\text{solute}}{\text{1}\text{liter}\text{of}\text{solution}}$

Molarity is estimation of moles in the total volume of the solution while molality is estimation of moles in relationship with solvent in the solution.

Mole fraction (X): Mole fraction is moles of each component is divided by total mass of the mixture.

${\text{X}}_{\text{solute}}\text{=}\frac{\text{moles}\text{of}\text{solute}}{\text{mass}\text{of}\text{the}\text{solution}}$

${\text{X}}_{\text{solvent}}\text{=}\frac{\text{moles}\text{of}\text{solvent}}{\text{mass}\text{of}\text{the}\text{solution}}$

Percent by mass: Mass percent is mass of the element is divided by total mass of the compound and multiplied by 100.

$\text{Percent mass =}\frac{\text{Mass}\text{of}\text{the}\text{element}}{\text{total}\text{mass}\text{of}\text{the}\text{compound}}\text{×100\%}$

To explain change in molarity