Chapter 4, Problem 4.46QE

(a)

Interpretation Introduction

Interpretation:

The mass of solute in $\text{0}\text{.113}\text{L}\text{of}\text{1}\text{.0}\text{M}\text{KBr}$ 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 4.46QE

The mass of solute in $\text{0}\text{.113}\text{L}\text{of}\text{1}\text{.0}\text{M}\text{KBr}$ solution is $13.45g.$

Explanation of Solution

Given: Volume of solution is $\text{0}\text{.113}\text{L}$ ; Molarity is $\text{1}\text{.0}\text{M}$

By substituting the given data into the molarity formula,

  $\begin{array}{c}\text{Molarity}\text{=}\frac{\text{Moles}\text{of}\text{solute}}{\text{Volume of solution in L}}\\ \\ \text{Moles}\text{of}\text{solute}\text{=}\left(\text{1}\text{.0}\text{M}\right)\left(\text{0}\text{.113L}\right)\text{=}0.11\text{3}\text{mole}\text{.}\end{array}$

Conversion of moles into mass:

  $\begin{array}{c}\text{Mass}\text{of}\text{solute}\text{=}\left(\text{moles}\right)\left(\text{Molar}\text{mass}\text{of}\text{KBr}\right)\\ \\ \text{=}\left(\text{0}\text{.113}\text{mole}\right)\left(119.002\text{g/mol}\right)\\ \\ \text{=}13.45g.\end{array}$

Therefore, the mass of solute solution is $13.45g.$

(b)

Interpretation Introduction

Interpretation:

The mass of solute in $\text{120}\text{mL}\text{of}\text{2}\text{.11}\text{M}{\text{KNO}}_3$ 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 4.46QE

The mass of solute in $\text{120}\text{mL}\text{of}\text{2}\text{.11}\text{M}{\text{KNO}}_3$ solution is $25.6g.$

Explanation of Solution

Given: Volume of solution is $\text{120}\text{mL}\text{=}\text{0}\text{.12}\text{L}$ ; Molarity is $\text{2}\text{.11}\text{M}\text{.}$

By substituting the given data into the molarity formula,

  $\begin{array}{c}\text{Molarity}\text{=}\frac{\text{Moles}\text{of}\text{solute}}{\text{Volume of solution in L}}\\ \\ \text{Moles}\text{of}\text{solute}\text{=}\left(\text{2}\text{.11}\text{M}\right)\left(\text{0}\text{.120}\text{L}\right)\text{=}0.2532\text{mole}\text{.}\end{array}$

Conversion of moles into mass:

  $\begin{array}{c}\text{Mass}\text{of}\text{solute}\text{=}\left(\text{moles}\right)\left(\text{Molar}\text{mass}\text{of}{\text{KNO}}_{\text{3}}\right)\\ \\ =\left(0.2532\text{mole}\right)\left(101.10\text{g/mol}\right)\\ \\ =25.6g.\end{array}$

Therefore, the mass of solute in $\text{120}\text{mL}\text{of}\text{2}\text{.11}\text{M}{\text{KNO}}_3$ solution is $25.6g.$