Chapter 17, Problem 22E

(a)

Interpretation Introduction

Interpretation:

The mass of KCl required to get 100 mL of 2.0m of KCl in water needs to be determined, if density of water is 1.00 g/cm³

Concept Introduction:

The measure of concentration of solute in solution or the amount of substance that is present in a specified amount of mass of solvent is termed as Molality.

Molarity is the number of moles of solute per liter of solution. Most of the solution concentration is indicated in terms of Molarity as it will indicate the volume of solvent or amount of solute.

Answer to Problem 22E

For preparing 100 mL of 2.0 m KCl solution, it is necessary to dissolve 15 g in 100 ml water.

Explanation of Solution

Given density of water as 1.00 g/cm³

Volume of water 100 mL

  $\begin{array}{l}\text{Density=}\frac{\text{Mass}}{\text{Volume}}\\ \text{Mass = Density x Volume}\\ \text{ = 1 }\frac{\text{g}}{c{m}^3}\text{x 100ml=100 g}\end{array}$

To convert to kg, divide by 1000

  $\text{=}\frac{\text{100}}{1000}kg=0.1kg$

Given molality of KCl as 2 m and to get molar mass,

  $\text{Molar mass of KCl is (39}\text{.10g + 35}\text{.45g) = 74}\text{.55g}$

As Molality is 2 m, it indicates 2 mol KCl in 1 kg of water i.e. $\frac{\text{2 mol KCl}}{{\text{1 kg of H}}_2\text{O}}$

Molarity is 74.55 g is present in 1 mol KCl.

To compute the mass of KCl in 2m, multiply both with mass of solvent in kg

  $\frac{\text{2 mol KCl}}{{\text{1 kg of H}}_2\text{O}}\text{x}\frac{\text{74}\text{.55g KCl}}{\text{1 mol KCl}}\text{x0}{\text{.100 kg of H}}_{2}O=1\text{4}\text{.91g of KCl}\approx \text{15g KCl}$

Therefore, to prepare 100 ml of 2m KCl solution, we require 15g of KCl in 100 mL of water.

(b)

Interpretation Introduction

Interpretation:

The mass of NaOH required to get 100 mL of 15% of NaOH in water needs to be determined, if density of water is 1.00 g/cm³

Concept Introduction:

The measure of concentration of solute in solution or the amount of substance that is present in a specified amount of mass of solvent is termed as Molality.

Molarity is the number of moles of solute per liter of solution. Most of the solution concentration is indicated in terms of Molarity as it will indicate the volume of solvent or amount of solute.

Answer to Problem 22E

To prepare 100 mL of 15% of NaOH in water, we require 18 grams of NaOH.

Explanation of Solution

Given density of water as 1.00 g/cm³

Volume of water 100 mL

  $\begin{array}{l}\text{Density=}\frac{\text{Mass}}{\text{Volume}}\\ \text{Mass = Density x Volume}\\ \text{ = 1 }\frac{\text{g}}{c{m}^3}\text{x 100ml=100 g}\end{array}$

Give mass percentage of NaOH as 15% which means 100g of solution contains 15 grams of NaOH solute.

Mass of NaOH solute =15g

Mass of solution = 100g

Hence, Mass of solution = Mass of NaOH solute + Mass of solvent

OR

Mass of solvent = Mass of solution − Mass of NaOH solute

  = 100 g - 15g

  = 85 g

In other words to prepare 15% NaOH solution by mass, 15g of NaOH solute is dissolved in 85 g of solvent.

In order to find out the mass of NaOH required to prepare 15% solution in 100 ml solvent, the mass of water is multiplied by mass percentage conversion factor:

  ${\text{100 g H}}_{\text{2}}\text{O x }\frac{15g\text{ NaOH}}{85g\text{ NaOH}}=17.6g\text{ NaOH}\approx \text{18 g NaOH}$

Therefore, to prepare 100 mL of 15% NaOH solution, we need to dissolve 18g of NaOH in 100mL of water.

(c)

Interpretation Introduction

Interpretation:

The mass of NaOH required to get 100 mL of 25% of NaOH in CH₃OH needs to be determined, if density of CH₃OH is 0.79g/cm³.

Concept Introduction:

The measure of concentration of solute in solution or the amount of substance that is present in a specified amount of mass of solvent is termed as Molality.

Molarity is the number of moles of solute per liter of solution. Most of the solution concentration is indicated in terms of Molarity as it will indicate the volume of solvent or amount of solute.

Answer to Problem 22E

To prepare 100 mL of 25% of NaOH in CH₃OH, we require 26 grams of NaOH.

Explanation of Solution

Given density of CH₃OH as 0.79 g/cm³

Volume of CH₃OH 100 mL

  $\begin{array}{l}\text{Density=}\frac{\text{Mass}}{\text{Volume}}\\ \text{Mass = Density x Volume}\\ \text{ = 0}\text{.79 }\frac{\text{g}}{c{m}^3}\text{x 100ml=79 g}\end{array}$

Given the mass percent of NaOH is 25% which indicates that 100 g of solution contains 25 g of NaOH solute.

Mass of NaOH solute = 25 g

Mass of solution = 100 g

Hence, Mass of solution = Mass of NaOH solute + Mass of solvent

OR

Mass of solvent = Mass of solution − Mass of NaOH solute

  = 100 g - 25g

  = 75 g

In other words to prepare 25% NaOH solution by mass, 25g of NaOH solute is dissolved in 75 g of solvent.

In order to find out the mass of NaOH required to prepare 25% solution in 100 ml solvent, the mass of CH₃OH is multiplied by mass percentage conversion factor:

  ${\text{100 g H}}_{\text{2}}\text{O x }\frac{15g\text{ NaOH}}{85g\text{ NaOH}}=17.6g\text{ NaOH}\approx \text{18 g NaOH}$

Therefore, to prepare 100 mL of 15% NaOH solution, we need to dissolve 18g of NaOH in 100mL of water.

  ${\text{79g CH}}_3\text{OH x }\frac{25g\text{ NaOH}}{75g{\text{ CH}}_3\text{OH}}=26.3g\text{ NaOH}\approx \text{26g NaOH}$

Therefore, to prepare 100 mL of 25% NaOH solution, we need to dissolve 26g of CH₃OH in 100mL of water.

(d)

Interpretation Introduction

Interpretation:

The mass of ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}$ required to get 100 mL of 0.10 mole of ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}$ in water needs to be determined.

Concept Introduction:

The measure of concentration of solute in solution or the amount of substance that is present in a specified amount of mass of solvent is termed as Molality.

Molarity is the number of moles of solute per liter of solution. Most of the solution concentration is indicated in terms of Molarity as it will indicate the volume of solvent or amount of solute.

Mole fraction is the number of molecules of particular component in a mixture divided by total number of moles

Answer to Problem 22E

To prepare 100 ml of 0.10 mole of ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}$ in water, we require 111g of ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}$ .

Explanation of Solution

Given density of water as 1.00 g/cm³

Volume of water 100 ml

  ${\text{Molar mass of H}}_2\text{O is (2x1}\text{.008g + 16g) = 18}\text{.02g}$

  $\begin{array}{l}\text{Density=}\frac{\text{Mass}}{\text{Volume}}\\ \text{Mass = Density x Volume}\\ \text{ = 1}\frac{\text{g}}{c{m}^3}\text{x 100ml=100g}\end{array}$

To compute the number of moles of water required

  $100\text{ x }\frac{1{\text{mol H}}_2\text{O}}{18g}=5.55{\text{mol H}}_2\text{O}$

To find the mole fraction of ${\text{ C}}_6{\text{H}}_{12}{\text{O}}_6$ , the following steps are taken

  $\begin{array}{l}{\text{Mole fraction of C}}_6{\text{H}}_{12}{\text{O}}_6=\frac{\text{no}{\text{.of moles of C}}_6{\text{H}}_{12}{\text{O}}_6}{\text{no}{\text{.of moles of C}}_6{\text{H}}_{12}{\text{O}}_6+\text{no}{\text{.of moles of H}}_2\text{O}}\\ \text{ = }\frac{\text{0}{\text{.10 mol C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}}{\text{1mol solution}}\\ \end{array}$

Hence, mole fraction of ${\text{C}}_6{\text{H}}_{12}{\text{O}}_6$ is 0.10 mol

To get the number of moles of water,

  $\begin{array}{l}\text{No}{\text{.of moles of H}}_2\text{O solvent= no}{\text{. of moles of solution - no of moles of C}}_6{\text{H}}_{12}{\text{O}}_6\\ \text{ =1}\text{.00 mol - 0}\text{.10 mol}\\ \text{ = 0}\text{.90 mol}\end{array}$

Mole fraction of ${\text{C}}_6{\text{H}}_{12}{\text{O}}_6$ is 0.10 mol which means that 0.10 moles of ${\text{C}}_6{\text{H}}_{12}{\text{O}}_6$ is dissolved in 0.90 mole of water.

Also, we found that no. of moles of water present in 100 ml is 5.55 mol

Molar mass of ${\text{C}}_6{\text{H}}_{12}{\text{O}}_6$ = {(6 x 12) + (12 x 1) + (6 x 16)} =180 g

To compute the mass of ${\text{C}}_6{\text{H}}_{12}{\text{O}}_6$ dissolved in solvent, multiply the number of moles of water and conversion factor obtained mole fraction and molar mass.

  $\text{5}{\text{.55 mol H}}_2\text{O x}\left[\frac{\text{0}{\text{.10 mol C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}}{\text{0}{\text{.9mol H}}_2\text{O}}\right]\text{x}\left[\frac{{\text{180g of C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}}{{\text{1mol C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}}\right]{\text{= 111g of C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}$

Therefore, to prepare 100 ml of 0.10 mole of ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}$ , we require 111g of ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}$ in 100 ml of water.