Chapter 13, Problem 46QRT

(a)

Interpretation Introduction

Interpretation:

The molarity of the solute $\text{14}\text{.2g}\text{KCl in 250mL}$ solution has to be calculated.

Concept introduction:

Molarity (M): Molarity is number of moles of the solute present in the one liter of the solution.

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

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

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

Moles: One mole is equivalent to the mass of the substance consists of same number of units equal to the atoms present in $\text{12}\text{g}$ of ${}^{\text{12}}\text{C}$.

Answer to Problem 46QRT

The molarity obtained as $0.762M.$

Explanation of Solution

Given mass is $\text{14}\text{.2}\text{g}$; Volume is $\text{250 mL}$; Molar mass $\text{KCl is 74}\text{.55 g/mol}.$

Calculate the number of moles:

  $\begin{array}{c}\text{Number of moles = }\frac{\text{Mass}}{\text{Molar}\text{mass}}\\ \\ \text{=}\frac{\text{14}\text{.2}\text{g}}{\text{74}\text{.55}\text{g/mol}}\\ \\ \text{=}\text{190}\text{.5×}{\text{10}}^{\text{-3}}\text{moles}\text{.}\end{array}$

Calculate the molarity:

  $\begin{array}{c}\text{ Molarity}\text{ =}\frac{\text{Number of moles}}{\text{volume}}\\ \\ \text{=}\frac{\text{190}\text{.5×}{\text{10}}^{\text{-3}}\text{moles}}{\text{250}\text{ml}}\\ \\ \text{=}0.762M\text{.}\end{array}$

Therefore, themolarity of the solute $\text{14}\text{.2g}\text{KCl in 250mL}$ solution was $0.762M.$

(b)

Interpretation Introduction

Interpretation:

The molarity of the solute $\text{5}\text{.08g}{\text{K}}_2{\text{CrO}}_4\text{ in 150mL}$ solution has to be calculated.

Concept introduction:

Refer part (a).

Answer to Problem 46QRT

The molarity obtained as $0.1744M.$

Explanation of Solution

Given mass is $\text{5}\text{.08}\text{g}$; Volume is $1\text{50 mL}$; Molar mass ${\text{K}}_{\text{2}}{\text{CrO}}_{\text{4}}\text{is}\text{194}\text{.19 g/mol}\text{.}$

Calculate the number of moles:

  $\begin{array}{c}\text{Number of moles = }\frac{\text{Mass}}{\text{Molar}\text{mass}}\\ \\ \text{=}\frac{\text{5}\text{.08}\text{g}}{\text{194}\text{.19}\text{g/mol}}\\ \\ \text{=}\text{26}\text{.16×}{\text{10}}^{\text{-3}}\text{moles}\text{.}\end{array}$

Calculate the molarity:

  $\begin{array}{c}\text{ Molarity}\text{ =}\frac{\text{Number of moles}}{\text{volume}}\\ \\ \text{=}\frac{\text{26}\text{.16×}{\text{10}}^{\text{-3}}\text{moles}}{\text{150}\text{ml}}\\ \\ \text{=}0.1744M\text{.}\end{array}$

Therefore, themolarity of the solute $\text{5}\text{.08g}{\text{K}}_2{\text{CrO}}_4\text{ in 150mL}$ solution was $0.1744M.$

(c)

Interpretation Introduction

Interpretation:

The molarity of the solute $\text{0}\text{.799g}{\text{KMnO}}_{\text{4}}\text{ in 400 mL}$ solution has to be calculated.

Concept introduction:

Refer part (a).

Answer to Problem 46QRT

The molarity obtained as $0.01265M\text{.}$

Explanation of Solution

Given mass is $\text{0}\text{.799}\text{g}$; Volume is $400\text{ mL}$; Molar mass ${\text{KMnO}}_{\text{4}}\text{is}\text{158}\text{.034 g/mol}\text{.}$

Calculate the number of moles:

  $\begin{array}{c}\text{Number of moles = }\frac{\text{Mass}}{\text{Molar}\text{mass}}\\ \\ \text{=}\frac{0.799\text{g}}{\text{158}\text{.034 g/mol}}\\ \\ \text{=}\text{5}\text{.06×}{\text{10}}^{\text{-3}}\text{moles}\text{.}\end{array}$

Calculate the molarity:

  $\begin{array}{c}\text{ Molarity}\text{ =}\frac{\text{Number of moles}}{\text{volume}}\\ \\ \text{=}\frac{\text{5}\text{.06×}{\text{10}}^{\text{-3}}\text{moles}}{\text{400}\text{ml}}\\ \\ \text{=}0.01265M\text{.}\end{array}$

Therefore, themolarity of the solute $\text{0}\text{.799g}{\text{KMnO}}_{\text{4}}\text{ in 400 mL}$ solution was $0.01265M\text{.}$

(d)

Interpretation Introduction

Interpretation:

The molarity of the solute $\text{15}\text{.0g}{\text{C}}_2{\text{H}}_{12}{\text{O}}_6\text{ in 500mL}$ solution has to be calculated.

Concept introduction:

Refer part (a).

Answer to Problem 46QRT

The molarity obtained as $0.227M$.

Explanation of Solution

Given mass is $\text{15}\text{.0}\text{g}$; Volume is $50\text{0 mL}$; Molar mass ${\text{C}}_{\text{2}}{\text{H}}_{12}{\text{O}}_6\text{is}\text{132}\text{.11 g/mol}\text{.}$

Calculate the number of moles:

  $\begin{array}{c}\text{Number of moles = }\frac{\text{Mass}}{\text{Molar}\text{mass}}\\ \\ \text{=}\frac{\text{15}\text{.0}\text{g}}{\text{132}\text{.11 g/mol}}\\ \\ \text{=}\text{113}\text{.5×}{\text{10}}^{\text{-3}}\text{moles}\text{.}\end{array}$

Calculate the molarity:

  $\begin{array}{c}\text{ Molarity}\text{ =}\frac{\text{Number of moles}}{\text{volume}}\\ \\ \text{=}\frac{\text{113}\text{.5×}{\text{10}}^{\text{-3}}\text{moles}}{\text{500}\text{ml}}\\ \\ \text{=}0.227M\text{.}\end{array}$

Therefore, the molarity of the solute $\text{15}\text{.0g}{\text{C}}_2{\text{H}}_{12}{\text{O}}_6\text{ in 500mL}$ solution was $0.227M.$