Chapter 13, Problem 13.14QP

Interpretation Introduction

Interpretation: The conversion between molarity, molality and percent by mass to be outlined.

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.

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

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

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