Chapter 12, Problem 12.19QE

(a)

Interpretation Introduction

Interpretation:

The concentration of ${\text{[Ca(NO}}_{\text{3}}{\text{)}}_{\text{2}}\text{]}$ in mass percentage has to be calculated.

Concept Introduction:

Mass percentage:  The mass percentage gives the number of grams of solute present in 100g of solution.  The formula of mass percentage is,

  $\text{Mass}\text{percentage=}\frac{\text{Grams}\text{of}\text{solute}}{\text{Grams}\text{of}\text{solution}}\text{×100\%}$

Answer to Problem 12.19QE

The concentration of ${\text{[Ca(NO}}_{\text{3}}{\text{)}}_{\text{2}}\text{]}$ in mass percentage is $\text{1}\text{.04\%}\text{.}$

Explanation of Solution

Given,

Mass of ${\text{[Ca(NO}}_{\text{3}}{\text{)}}_{\text{2}}\text{]}$    =$\text{4}\text{.50}\text{g}$

Mass of water        =$\text{430}\text{.0g}$

Mass of solution    =$\text{434}\text{.5g}$

The mass percentage of ${\text{[Ca(NO}}_{\text{3}}{\text{)}}_{\text{2}}\text{]}$ is given as,

  $\begin{array}{l}\text{Mass}\text{percentage=}\frac{\text{Grams}\text{of}\text{solute}}{\text{Grams}\text{of}\text{solution}}\text{×100\%}\\ \\ \text{Mass}\text{percentage=}\frac{\text{4}\text{.5g}}{\text{434}\text{.5}\text{g}}\text{×100\%}\\ \\ \text{Mass}\text{percentage}\text{=1}\text{.0356\%}\approx \text{1}\text{.04\%}\end{array}$

The concentration of ${\text{[Ca(NO}}_{\text{3}}{\text{)}}_{\text{2}}\text{]}$ in mass percentage is $\text{1}\text{.04\%}\text{.}$

(b)

Interpretation Introduction

Interpretation:

The concentration of ${\text{[Ca(NO}}_{\text{3}}{\text{)}}_{\text{2}}\text{]}$ in mole fraction has to be calculated.

Concept Introduction:

Mole fraction:  Mole fraction of a substance in a solution is the number of moles of that substance divided by the total number of moles of all substances present.  The formula is,

  $\begin{array}{l}{\text{χ}}_{\text{A}}\text{=}\frac{\text{Moles}\text{of}\text{A (in}\text{mol)}}{\text{Moles}\text{of}\text{A (in mol) +Moles}\text{of}\text{B (in mol) +Moles}\text{of}\text{C (in mol) +}\mathrm{...}}\\ \\ {\text{χ}}_{\text{A}}\text{=}\frac{\text{Moles}\text{of}\text{A (in}\text{mol)}}{\text{Total}\text{number}\text{of}\text{moles}\text{of}\text{components}\text{(in}\text{mol)}}\end{array}$

Answer to Problem 12.19QE

The concentration of ${\text{[Ca(NO}}_{\text{3}}{\text{)}}_{\text{2}}\text{]}$ in mole fraction is $\text{0}\text{.00114}\text{.}$

Explanation of Solution

Given,

Mass of ${\text{[Ca(NO}}_{\text{3}}{\text{)}}_{\text{2}}\text{]}$    =$\text{4}\text{.50}\text{g}$

Mass of water        =$\text{430}\text{.0g}$

Mass of solution    =$\text{434}\text{.5g}$

The moles of ${\text{[Ca(NO}}_{\text{3}}{\text{)}}_{\text{2}}\text{]}$=$\text{4}\text{.50g×}\frac{\text{1}\text{mol}}{\text{164}\text{.088}\text{g}}\text{=0}\text{.0274}\text{moles}$

The moles of water=$\text{430g×}\frac{\text{1}\text{mol}}{\text{18}\text{.02}\text{g}}\text{=23}\text{.8623moles}$

The mole fraction of ${\text{[Ca(NO}}_{\text{3}}{\text{)}}_{\text{2}}\text{]}$,

  $\begin{array}{l}{\text{χ}}_{{}_{{\text{Ca(NO}}_{\text{3}}{\text{)}}_{\text{2}}}}\text{=}\frac{\text{Moles}\text{of}{\text{[Ca(NO}}_{\text{3}}{\text{)}}_{\text{2}}\text{] (in}\text{mol)}}{\text{Total}\text{number}\text{of}\text{moles}\text{of}\text{components}\text{(in}\text{mol)}}\\ \\ {\text{χ}}_{{\text{Ca(NO}}_{\text{3}}{\text{)}}_{\text{2}}}\text{=}\frac{\text{0}\text{.0274}\text{moles}}{\text{0}\text{.0274}\text{moles+ 23}\text{.8623}\text{moles}}\\ \\ {\text{χ}}_{{\text{Ca(NO}}_{\text{3}}{\text{)}}_{\text{2}}}\text{=}\frac{\text{0}\text{.0274}\text{moles}}{\text{23}\text{.8897}\text{moles}}\\ \\ {\text{χ}}_{{\text{Ca(NO}}_{\text{3}}{\text{)}}_{\text{2}}}\text{=0}\text{.00114}\end{array}$

The concentration of ${\text{[Ca(NO}}_{\text{3}}{\text{)}}_{\text{2}}\text{]}$ in mole fraction is $\text{0}\text{.00114}\text{.}$

(c)

Interpretation Introduction

Interpretation:

The concentration of ${\text{[Ca(NO}}_{\text{3}}{\text{)}}_{\text{2}}\text{]}$ in molality has to be calculated.

Concept Introduction:

Molality:  Molality is defined as moles of solute per specified amount of mass of the solvent.  The formula is,

  $\text{Molality}\text{(m)=}\frac{\text{Mass of}\text{solute (in mol)}}{\text{Mass}\text{of}\text{solvent (in kg)}}$

Answer to Problem 12.19QE

The concentration of ${\text{[Ca(NO}}_{\text{3}}{\text{)}}_{\text{2}}\text{]}$ in molality is $\text{0}\text{.64m}\text{.}$

Explanation of Solution

Given,

Mass of ${\text{[Ca(NO}}_{\text{3}}{\text{)}}_{\text{2}}\text{]}$    =$\text{25}\text{.0}\text{g}$

Mass of water        =$\text{430}\text{.0g=0}\text{.43 kg}$

Mass of solution    =$\text{525}\text{.0g}$

The moles of ${\text{[Ca(NO}}_{\text{3}}{\text{)}}_{\text{2}}\text{]}$=$\text{4}\text{.50g×}\frac{\text{1}\text{mol}}{\text{164}\text{.088}\text{g}}\text{=0}\text{.0274}\text{moles}$

The concentration of ${\text{[Ca(NO}}_{\text{3}}{\text{)}}_{\text{2}}\text{]}$ is,

  $\begin{array}{l}\text{Molality}\text{(m)=}\frac{\text{Mass of}\text{solute (in mol)}}{\text{Mass}\text{of}\text{solvent (in kg)}}\\ \\ \text{m=}\frac{\text{0}\text{.0274}\text{moles}}{\text{0}\text{.43kg}}\\ \\ \text{m=0}\text{.0637}\text{m}\approx \text{0}\text{.064}\text{m}\end{array}$

The concentration of ${\text{[Ca(NO}}_{\text{3}}{\text{)}}_{\text{2}}\text{]}$ in molality is $\text{0}\text{.64m}\text{.}$