Chapter 4, Problem 4.109QP

(a)

Interpretation Introduction

Interpretation:

Molar concentration of ${\text{Cl}}^{\text{-}}$ ion present in given ${\text{CaCl}}_{\text{2}}$ solution and gram mass of ${\text{Cl}}^{\text{-}}$ ion present in $\text{0}\text{.100 L}$ given ${\text{CaCl}}_{\text{2}}$ solution should be calculated.

Concept introduction:

Molarity:

The concentration of the solutions is given by the term of molarity and it is given by ratio between numbers of moles of solute present in litter of solution.

$\text{Molarity}\text{=}\frac{\text{No}\text{.mole}}{\text{volume}\text{(L)}}$

Mole:

The mole of compound is given by the ratio between taken mass of the compound and molar mass of the compound.

$\text{Mole}\text{=}\frac{\text{Massof}\text{the}\text{compound}}{\text{Molar}\text{mass}\text{of}\text{the}\text{compound}}$

Mole:

The mole of the solute is calculated by multiplication of concentration of solution and volume of the solution and it is,

$\text{Mole}\text{=}\text{Concentration}\text{(M)×}\text{volume}\text{(L)}$

 (b)

Interpretation Introduction

Interpretation:

Molar concentration of ${\text{Cl}}^{\text{-}}$ ion present in given ${\text{CaCl}}_{\text{2}}$ solution and gram mass of ${\text{Cl}}^{\text{-}}$ ion present in $\text{0}\text{.100 L}$ given ${\text{CaCl}}_{\text{2}}$ solution should be calculated.

Concept introduction:

Molarity:

The concentration of the solutions is given by the term of molarity and it is given by ratio between numbers of moles of solute present in litter of solution.

$\text{Molarity}\text{=}\frac{\text{No}\text{.mole}}{\text{volume}\text{(L)}}$

Mole:

The mole of compound is given by the ratio between taken mass of the compound and molar mass of the compound.

$\text{Mole}\text{=}\frac{\text{Massof}\text{the}\text{compound}}{\text{Molar}\text{mass}\text{of}\text{the}\text{compound}}$

Mole:

The mole of the solute is calculated by multiplication of concentration of solution and volume of the solution and it is,

$\text{Mole}\text{=}\text{Concentration}\text{(M)×}\text{volume}\text{(L)}$