Chapter 4, Problem 4.45QE

(a)

Interpretation Introduction

Interpretation:

The mass of solute in $\text{3}\text{.13L}\text{of}\text{2}\text{.21M}\text{HCl}$ solution has to be calculated.

Concept introduction:

Molarity: Dividing moles of solute by the volume of solution in Liter.

$\text{Molarity}\text{=}\frac{\text{Moles}\text{of}\text{solute}}{\text{Volume of solution in L}}$

Answer to Problem 4.45QE

The mass of solute in $\text{3}\text{.13L}\text{of}\text{2}\text{.21M}\text{HCl}$ solution is $252g.$

Explanation of Solution

Given: Volume of solution is $\text{3}\text{.13}\text{L}$ ; Molarity is $\text{2}\text{.21}\text{M}$

By substituting the given data into the molarity formula,

$\begin{array}{c}\text{Molarity}\text{=}\frac{\text{Moles}\text{of}\text{solute}}{\text{Volume of solution in L}}\\ \\ \text{Moles}\text{of}\text{solute}\text{=}\left(\text{2}\text{.21M}\right)\left(\text{3}\text{.13L}\right)\text{=}\text{6}\text{.9173}\text{mole}\text{.}\end{array}$

Conversion of moles into mass:

$\begin{array}{c}\text{Mass}\text{of}\text{solute}\text{=}\left(\text{moles}\right)\left(\text{Molar}\text{mass}\text{of}\text{HCl}\right)\\ \\ =\left(\text{6}\text{.9173}\text{mole}\right)\left(\text{36}\text{.5}\text{g/mol}\right)\\ \\ =252g.\end{array}$

Therefore, the mass of solute solution is $252g.$

(b)

Interpretation Introduction

Interpretation:

The mass of solute in $\text{1}\text{.5L}\text{of}\text{1}\text{.2M}\text{KCl}$ solution has to be calculated.

Concept introduction:

Molarity: Dividing moles of solute by the volume of solution in Liter.

  $\text{Molarity}\text{=}\frac{\text{Moles}\text{of}\text{solute}}{\text{Volume of solution in L}}$

Answer to Problem 4.45QE

The mass of solute in $\text{1}\text{.5L}\text{of}\text{1}\text{.2M}\text{KCl}$ solution is $134g.$

Explanation of Solution

Given: Volume of solution is $\text{1}\text{.5}\text{L}$ ; Molarity is $\text{1}\text{.2}\text{M}\text{.}$

By substituting the given data into the molarity formula,

  $\begin{array}{c}\text{Molarity}\text{=}\frac{\text{Moles}\text{of}\text{solute}}{\text{Volume of solution in L}}\\ \\ \text{Moles}\text{of}\text{solute}\text{=}\left(\text{1}\text{.2}\text{M}\right)\left(\text{1}\text{.5}\text{L}\right)\text{=}1.8\text{mole}\text{.}\end{array}$

Conversion of moles into mass:

$\begin{array}{c}\text{Mass}\text{of}\text{solute}\text{=}\left(\text{moles}\right)\left(\text{Molar}\text{mass}\text{of}\text{KCl}\right)\\ \\ =\left(\text{1}\text{.8}\text{mole}\right)\left(\text{74}\text{.55}\text{g/mol}\right)\\ \\ =134g.\end{array}$

Therefore, the mass of solute in $\text{1}\text{.5L}\text{of}\text{1}\text{.2M}\text{KCl}$ solution is $134g.$