Chapter 13, Problem 13.19QP

(a)

Interpretation Introduction

Interpretation:

The molality of the given solution is to be determined.

Concept introduction:

Molality (m):

Molality is the number of moles of solute present in one kilogram of solvent.

$\text{Molality (m) =}\frac{\text{Number}\text{of}\text{moles}\text{of}\text{solute}}{\text{1kg}\text{of}\text{solvent}}$

Answer to Problem 13.19QP

The molality of the given solution is $1.74\text{m}$

Explanation of Solution

Given,

$\begin{array}{c}Strengthofsugarsolution\text{=}\text{1}\text{.22}\text{M}\\ Densityofsugarsolution\text{=}\text{1}\text{.12g /mL}\\ Molecularweightofsugar=\text{342}\text{.3g/mol}\end{array}$

Mass of sugar is calculated,

$\begin{array}{c}\text{mass of sugar =1}\text{.22}\text{mol}\times \frac{342.3\text{g}}{1mol}\\ \text{=}\text{417}\text{.6}\text{g}\\ \text{=}\text{0}\text{.4176}\text{kg}\end{array}$

$\begin{array}{c}\text{Mass of 1L solution=1000mL×}\frac{\text{1}\text{.12g}}{\text{1mL}}\\ \text{=}\text{1120}\text{g}\\ \text{=}\text{1}\text{.12}\text{kg}\end{array}$

$\begin{array}{c}\text{mass of solvent =}\left(1.12kg\right)-\left(\text{0}\text{.4176}\text{kg}\right)\\ \text{=}\text{0}\text{.7024}k\text{g}\end{array}$

Therefore,

The molality of the given solution is,

$\begin{array}{c}\text{Molality (m) =}\frac{\text{1}\text{.22}\text{mol}}{\text{0}\text{.7024}\text{kg}}\\ =1.74\text{m}\end{array}$

The molality of the given solution is $1.74\text{m}$

(b)

Interpretation Introduction

Interpretation:

The molality of the given solution is to be determined.

Concept introduction:

Molality (m):

Molality is the number of moles of solute present in one kilogram of solvent.

$\text{Molality (m) =}\frac{\text{Number}\text{of}\text{moles}\text{of}\text{solute}}{\text{1kg}\text{of}\text{solvent}}$

Answer to Problem 13.19QP

The molality of the given solution is $0.865\text{m}$

Explanation of Solution

Given,

$\begin{array}{c}StrengthofNaOHsolution\text{=}\text{0}\text{.87}\text{M}\\ DensityofNaOHsolution\text{=}\text{1}\text{.04g /mL}\\ MolecularNaOHofsugar=\text{40}\text{g}\end{array}$

Mass of $NaOH$ sugar is calculated,

$\begin{array}{c}\text{mass of }NaOH\text{ =0}\text{.87}\text{mol}\times \frac{40\text{g}}{1mol}\\ \text{=}\text{34}\text{.8}\text{g}\\ \text{=}\text{0}\text{.0348}\text{kg}\end{array}$

$\begin{array}{c}\text{Mass of 1L solution=1000mL×}\frac{\text{1}\text{.04g}}{\text{1mL}}\\ \text{=}\text{1040}\text{g}\\ \text{=}\text{1}\text{.04}\text{kg}\end{array}$

$\begin{array}{c}\text{mass of solvent =}\left(\text{1}\text{.04}\text{kg}\right)-\left(\text{0}\text{.0348}\text{kg}\right)\\ \text{=}\text{1}\text{.0052}k\text{g}\end{array}$

Therefore,

The molality of the given solution is,

$\begin{array}{c}\text{Molality (m) =}\frac{\text{0}\text{.87}\text{mol}}{\text{1}\text{.0052}k\text{g}}\\ =0.865\text{m}\end{array}$

The molality of the given solution is $0.865\text{m}$

(c)

Interpretation Introduction

Interpretation:

The molality of the given solution is to be determined.

Concept introduction:

Molality (m):

Molality is the number of moles of solute present in one kilogram of solvent.

$\text{Molality (m) =}\frac{\text{Number}\text{of}\text{moles}\text{of}\text{solute}}{\text{1kg}\text{of}\text{solvent}}$

Answer to Problem 13.19QP

The molality of the given solution is $1.09\text{m}$

Explanation of Solution

Given,

$\begin{array}{c}StrengthofNaHC{O}_{3}solution\text{=}\text{5}\text{.24}\text{M}\\ DensityofNaHC{O}_{3}solution\text{=}\text{1}\text{.19}\text{g/mL}\\ MolecularNaHC{O}_{3}ofsugar=\text{84}\text{.01}\text{g}\end{array}$

Mass of $NaHC{O}_3$$NaOH$ sugar is calculated,

$\begin{array}{c}\text{mass of }NaHC{O}_3\text{ =5}\text{.24}\text{mol}\times \frac{84.01\text{g}}{1mol}\\ \text{=}\text{440}\text{.2}\text{g}\\ \text{=}\text{0}\text{.4402}\text{kg}\end{array}$

$\begin{array}{c}\text{Mass of 1L solution=1000mL×}\frac{\text{5}\text{.24g}}{\text{1mL}}\\ \text{=}\text{5240}\text{g}\\ \text{=}5.24\text{kg}\end{array}$

$\begin{array}{c}\text{mass of solvent =}\left(5.24\text{kg}\right)-\left(\text{0}\text{.4402}\text{kg}\right)\\ \text{=}\text{4}\text{.7998}k\text{g}\end{array}$

Therefore,

The molality of the given solution is,

$\begin{array}{c}\text{Molality (m) =}\frac{\text{5}\text{.24}\text{mol}}{\text{4}\text{.7998}k\text{g}}\\ =1.09\text{m}\end{array}$

The molality of the given solution is $1.09\text{m}$.