Chapter 13, Problem 60GQ

Interpretation Introduction

Interpretation: Molality of $Ca{\left(N{O}_3\right)}_2$ and the total molality of the ions in the solution are to be determined.

Concept introduction:

Mole fraction: Amount of that component divided by the total amount of all of the components of the mixture $\left({\text{ n}}_{\text{A}}{\text{ + n}}_{\text{B}}{\text{ + n}}_{\text{C }}\text{+ }\mathrm{...}\right)$.

  ${\text{Mole fraction of A (χ}}_{\text{A}}\text{)= }\frac{{\text{n}}_{\text{A}}}{{\text{ n}}_{\text{A}}{\text{ + n}}_{\text{B}}{\text{ + n}}_{\text{C }}\text{+ }\mathrm{...}}$

Weight percent: The mass of one component divided by the total mass of the mixture, multiplied by $100\%$

  $\text{weight \% A = }\frac{\text{Mass of A}}{\text{ Mass of A + Mass of B + Mass of C + }\mathrm{...}}\text{×100\%}$

The number of moles of any substance can be determined using the equation

  $\text{Number}\text{of}\text{mole}\text{=}\frac{\text{Given}\text{mass}\text{of}\text{the}\text{substance}}{\text{Molar}\text{mass}}$

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.

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

Density is calculated by the equation,

  $\text{Density}\text{=}\frac{\text{mass}}{\text{volume}}$