Chapter 15, Problem 79QRT

(a)

Interpretation Introduction

Interpretation:

The mass percentage of chlorine in a sample has to be calculated.

Concept Introduction:

Molarity:

The concentration is given in the term of molarity, it is given as,

    $\text{Molarity}\text{=}\frac{\text{moles}}{\text{liter}}$

Mole:

Mole of compound that present in given amount of solid is calculated by below,

    $\text{Mole}\text{=}\frac{\text{Mass}}{\text{Molar}\text{mass}}$

Mass percentage:

The amount of the analyte that present in given sample is calculated by mass percentage formula, it is

    $\text{Mass}\text{\%}\text{=}\frac{\text{Mass}\text{of analyte}}{\text{Total}\text{mass}\text{of}\text{sample}}\text{×100}$

(b)

Interpretation Introduction

Interpretation:

The mass percentage of chlorine in a sample has to be calculated by using given ${\text{K}}_{\text{sp}}$values.

Concept Introduction:

Molarity:

The concentration is given in the term of molarity, it is given as,

    $\text{Molarity}\text{=}\frac{\text{moles}}{\text{liter}}$

Mole:

Mole of compound that present in given amount of solid is calculated by below,

    $\text{Mole}\text{=}\frac{\text{Mass}}{\text{Molar}\text{mass}}$

Solubility product constant:

The equilibrium constant of a more soluble ionic compound in water at the higher solubility is known as solubility product constant.  The equilibrium constant of more soluble ionic compound is given by ${\text{K}}_{\text{sp}}$ and it is expressed by product of multiplication of number of each ion present in the compound rise to the power of number respective ion present in the compound to give a maximum solubility of the compound.

    $\begin{array}{l}{\text{A}}_{\text{a}}{\text{B}}_{\text{b}}\underset{}{\rightleftharpoons }{\text{A}}^{\text{a}}{\text{B}}^{\text{b}}\\ \\ {\text{K}}_{\text{sp}}\text{=}{\text{a(A)}}^{\text{a}}{\text{b(B)}}^{\text{b}}\end{array}$