Chapter 17, Problem 16E

Interpretation Introduction

Interpretation: The molarity, molality and the mole fraction of the each common commercial acids and bases needs to be determined which are the aqueous solution with given properties as follows:

Reagents	Density $\left({\text{g/cm}}^{\text{3}}\right)$	Mass Percent of the solute
Hydrochloric acid	$1.19$	$38$
Nitric acid	$1.42$	$70$
Sulfuric acid	$1.84$	$95$
Acetic acid	$1.05$	$99$
Ammonia	0.90	$28$

Concept introduction: Molarity of a solution can be defined as the number of moles of solute present in 1 litre or ${\text{dm}}^{\text{3}}$ solution and is expressed in mol/litre.

$\text{M}=\frac{\text{Mass}\text{of}\text{moles}\text{of}\text{solute}}{\text{Volume}\text{of}\text{solution}\text{in}\text{litres}}$

Molality of a solution can be defined as the number of moles of solute that can be dissolved in $1000$ grams (or 1 kg) of the solvent and is expressed in mol/kg.

$\text{Molality}\left(\text{m}\right)=\frac{\text{Number}\text{of}\text{moles}\text{of}\text{solute}}{\text{Mass}\text{of}\text{solvent}\text{in}\text{kg}}$

Mole fraction $\left(\chi \right)$ of the constituents (i.e. solute as well as solvent) can be defined as the fraction obtained by dividing the number of moles of the ‘ a’ constituent (n_a) by the total number of moles (n_T) of all the constituents which are present in the solution.

${\chi }_a=\frac{\text{Number}\text{of}\text{moles}\text{of}a\left({\text{n}}_{\text{a}}\right)}{\text{Total}\text{number}\text{of}\text{moles}\left({\text{n}}_{\text{T}}\right)}$

Answer to Problem 16E

The molarity, molality and mole fraction of the each reagent is as follows:

Reagents	Molarity (M)	Molality (m)	Mole fraction
Hydrochloric acid	12.38	16.97	0.23
Nitric acid	15.77	37.03	0.4
Sulfuric acid	17.83	193.8	0.7772
Acetic acid	17.33	1650	0.967
Ammonia	14.82	23.42	0.2754

Explanation of Solution

1. Hydrochloric acid (HCl):

Density of HCl solution is $\text{1}\text{.19}{\text{g/cm}}^{\text{3}}$ .

Therefore, the density of $\text{1}\text{L}$ solution = $\text{1}\text{.19}\times {\text{10}}^3\text{g/L}$

And $38\%$ of HCl by mass is present which means $\text{38}\text{g}$ of HCl is present in $\text{100}\text{mL}$ solution. Therefore,

$\begin{array}{c}\text{1}\text{L}\text{of}\text{solution contains the amount of HCl }=\frac{38\text{g}\times \text{1190}\left(\text{1}\text{L}\right)}{100\text{mL}}\\ =452.2g\end{array}$

$\begin{array}{c}\text{Therefore,}\text{the}\text{amount}\text{of}\text{water}\text{present}\text{in}\text{the}\text{solution}=1190\text{g}-452.2\text{g}\\ =737.8\text{g}\end{array}$

$\begin{array}{c}\text{Molarity}\left(\text{M}\right)\text{of}\text{HCl}\text{solution}=\frac{\text{weight}}{\text{m}\text{.wt}}\times \frac{\text{1}}{\text{V}\left(\text{1L}\right)}\\ =\frac{452.2}{36.5}\times \frac{1}{1\left(\text{L}\right)}\\ =12.38\text{M}\end{array}$

$\begin{array}{c}\text{Molality}\left(m\right)\text{of}\text{HCl}\text{solution}=\frac{\text{weight}}{\text{m}\text{.wt}}\times \frac{\text{1}}{\text{solvent}\left(\text{in}\text{kg}\right)}\\ =\frac{452.2}{36.5}\times \frac{1}{0.73\text{kg}}\\ =16.97\text{m}\end{array}$

$\text{Mole}\text{fraction}=\frac{\text{moles}\text{of}\text{solute}}{\text{total}\text{moles}}$

Moles of solute (HCl) = 12.38

$\left[\therefore \frac{\text{weight}}{\text{m}\text{.wt}}=\text{moles}\right]$

$\text{Moles}\text{of}\text{solvent}\left({\text{H}}_{\text{2}}\text{O}\right)=\frac{737.89}{18\text{g}}=40.98\text{moles}$

$\therefore \text{Mole}\text{fraction}\text{of}\text{HCl}=\frac{\text{12}\text{.38}}{\text{40}\text{.98+12}\text{.38}}=0.23$

2. Nitric acid $\left({\text{HNO}}_{\text{3}}\right)$:

$\text{Density}\text{=1}\text{.42}{\text{g/cm}}^{\text{3}}$

$\text{Density}\text{of}\text{1}\text{L}\text{solution}=1.42\times {10}^3\text{g/L}$

$\text{Mass}\text{percent}\text{of}\text{solute}=70\%$

$\begin{array}{c}\text{For}\text{1}\text{L}\text{of}\text{solution,}\text{the}\text{amount}\text{of}{\text{HNO}}_{\text{3}}\text{present}=\frac{70\text{g}\times \text{1420}\text{g/L}}{100\text{mL}}\\ =994\text{g}\end{array}$

$\begin{array}{c}\text{Amount}\text{of}\text{water}\text{present}\text{in}\text{the}\text{solution}=1420\text{g}-994\text{g}\\ =426\text{g}\end{array}$

$\begin{array}{c}\text{Molarity}\left(\text{M}\right)\text{of}{\text{HNO}}_3\text{solution}=\frac{\text{weight}}{\text{m}\text{.wt}}\times \frac{\text{1}}{\text{V}\left(\text{in}\text{L}\right)}\\ =\frac{994}{63}\times \frac{1}{1}\\ =15.77\text{M}\end{array}$

$\begin{array}{c}\text{Molality}\left(m\right)\text{of}{\text{HNO}}_3\text{solution}=\frac{\text{weight}}{\text{m}\text{.wt}}\times \frac{\text{1}}{\text{solvent}\left(\text{in}\text{kg}\right)}\\ =\frac{994\text{g}}{63}\times \frac{1}{0.426\text{kg}}\\ =37.03\text{m}\end{array}$

$\text{Mole}\text{fraction}=\frac{\text{moles}\text{of}\text{solute}}{\text{total}\text{moles}}$

$\text{Moles}\text{of}\left({\text{HNO}}_{\text{3}}\right)=\frac{994}{63}=15.77\text{moles}$$\left[\therefore \frac{\text{weight}}{\text{m}\text{.wt}}=\text{moles}\right]$

$\text{Moles}\text{of}\text{solvent}\left({\text{H}}_{\text{2}}\text{O}\right)=\frac{426}{18\text{g}}=23.66\text{moles}$

$\therefore \text{Mole}\text{fraction}\text{of}{\text{HNO}}_3=\frac{\text{15}\text{.77}}{\text{15}\text{.77+23}\text{.66}}=0.4$

3. Sulfuric acid $\left({\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\right)$:

$\text{Density}=1.84{\text{g/cm}}^{\text{3}}$

$\text{Density}\text{of}\text{1}\text{L}\text{solution}=1.84\times {10}^3\text{g/L}$

$\text{Mass}\text{percent}\text{of}\text{solute}=95\%$

$\begin{array}{c}\text{For}\text{1}\text{L}\text{of}\text{solution,}\text{the}\text{amount}\text{of}{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\text{present}=\frac{95\text{g}\times \text{1840}\text{g/L}}{100\text{mL}}\\ =1748\text{g}\end{array}$

$\begin{array}{c}\text{Amount}\text{of}\text{water}\text{present}\text{in}\text{the}\text{solution}=1840\text{g}-1748\text{g}\\ =92\text{g}\end{array}$

$\begin{array}{c}\text{Molarity}\left(\text{M}\right)\text{of}\left({\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\right)\text{solution}=\frac{\text{weight}\text{of}\text{solute}}{\text{m}\text{.wt}\text{of}\text{solute}}\times \frac{\text{1}}{\text{V}\left(\text{in}\text{L}\right)}\\ =\frac{1748\text{g}}{98}\times \frac{1}{1\left(\text{L}\right)}\\ =17.83\text{M}\end{array}$

$\begin{array}{c}\text{Molality}\left(m\right)\text{of}{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\text{solution}=\frac{\text{weight}}{\text{m}\text{.wt}}\times \frac{\text{1}}{\text{solvent}\left(\text{in}\text{kg}\right)}\\ =\frac{1748\text{g}}{98}\times \frac{1}{0.092\text{kg}}\\ =193.8\text{m}\end{array}$

$\text{Mole}\text{fraction}=\frac{\text{moles}\text{of}\text{solute}}{\text{total}\text{moles}}$

$\text{Moles}\text{of}\text{solute}\left({\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\right)=\frac{1748}{98}=17.83\text{moles}$$\left[\therefore \frac{\text{weight}}{\text{m}\text{.wt}}=\text{moles}\right]$

$\text{Moles}\text{of}\text{solvent}\left({\text{H}}_{\text{2}}\text{O}\right)=\frac{92}{18\text{g}}=5.11\text{moles}$

$\therefore \text{Mole}\text{fraction}\text{of}{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}=\frac{\text{17}\text{.83}}{\text{5}\text{.11+17}\text{.83}}=0.7772$

4. Acetic acid $\left({\text{CH}}_{\text{3}}\text{COOH}\right)$:

$\text{Density}=1.05{\text{g/cm}}^{\text{3}}$

$\begin{array}{c}\text{Density}\text{of}\text{1}\text{L}\text{solution}=1.05\times {10}^3\text{g/L}\\ =\text{1050}\text{g/L}\end{array}$

$\text{Mass}\text{percent}\text{of}\text{solute}=99\%$

$\begin{array}{c}\text{For}\text{1}\text{L}\text{of}\text{solution,}\text{the}\text{amount}\text{of}{\text{CH}}_{\text{3}}\text{COOH}\text{present}=\frac{99\times \text{1050}}{100}\\ =1039.5\text{g}\end{array}$

$\begin{array}{c}\text{Amount}\text{of}\text{water}\text{present}\text{in}\text{the}\text{solution}=1050\text{g}-1039.5\text{g}\\ =10.5\text{g}\end{array}$

$\begin{array}{c}\text{Molarity}\left(\text{M}\right)\text{of}{\text{CH}}_{\text{3}}\text{COOH}\text{solution}=\frac{\text{weight}\text{of}\text{solute}}{\text{m}\text{.wt}\text{of}\text{solute}}\times \frac{\text{1}}{\text{V}\left(\text{in}\text{L}\right)}\\ =\frac{1039.5\text{g}}{60}\times \frac{1}{1\left(\text{L}\right)}\\ =17.33\text{M}\end{array}$

$\begin{array}{c}\text{Molality}\left(m\right)\text{of}{\text{CH}}_{\text{3}}\text{COOH}\text{solution}=\frac{\text{weight}}{\text{m}\text{.wt}}\times \frac{\text{1}}{\text{solvent}\left(\text{in}\text{kg}\right)}\\ =\frac{1039.5\text{g}}{60}\times \frac{1}{0.0105\text{kg}}\\ =1650\text{m}\end{array}$

$\text{Mole}\text{fraction}=\frac{\text{moles}\text{of}\text{solute}}{\text{total}\text{moles}}$

$\text{Moles}\text{of}\text{solute}\left({\text{CH}}_{\text{3}}\text{COOH}\right)=\frac{1039.5}{60}=17.325\text{moles}$$\left[\therefore \frac{\text{weight}}{\text{m}\text{.wt}}=\text{moles}\right]$

$\text{Moles}\text{of}\text{solvent}\left({\text{H}}_{\text{2}}\text{O}\right)=\frac{10.5}{18\text{g}}=0.583\text{moles}$

$\therefore \text{Mole}\text{fraction}\text{of}{\text{CH}}_{\text{3}}\text{COOH}=\frac{\text{17}\text{.325}}{\text{17}\text{.325+0}\text{.583}}=0.967$

5. Ammonia $\left({\text{NH}}_{\text{3}}\right)$:

$\text{Density}=0.90{\text{g/cm}}^{\text{3}}$

$\begin{array}{c}\text{Density}\text{of}\text{1}\text{L}\text{solution}=0.90\times {10}^3\text{g/L}\\ =900\text{g/L}\end{array}$

$\text{Mass}\text{percent}\text{of}\text{solute}=28\%$

$\begin{array}{c}\text{For}\text{1}\text{L}\text{of}\text{solution,}\text{the}\text{amount}\text{of}{\text{NH}}_{\text{3}}\text{present}=\frac{28\times 900}{100}\\ =252\text{g}\end{array}$

$\begin{array}{c}\text{Amount}\text{of}\text{water}\text{present}\text{in}\text{the}\text{solution}=900\text{g}-252\text{g}\\ =648\text{g}\end{array}$

$\begin{array}{c}\text{Molarity}\left(\text{M}\right)\text{of}{\text{NH}}_{\text{3}}\text{solution}=\frac{\text{weight}\text{of}\text{solute}}{\text{m}\text{.wt}\text{of}\text{solute}}\times \frac{\text{1}}{\text{V}\left(\text{in}\text{L}\right)}\\ =\frac{252\text{g}}{17}\times \frac{1}{1\left(\text{L}\right)}\\ =14.82\text{M}\end{array}$

$\begin{array}{c}\text{Molality}\left(m\right)\text{of}{\text{NH}}_{\text{3}}\text{solution}=\frac{\text{weight}}{\text{m}\text{.wt}}\times \frac{\text{1}}{\text{solvent}\left(\text{in}\text{kg}\right)}\\ =\frac{252\text{g}}{17}\times \frac{1}{0.648\text{kg}}\\ =23.42\text{m}\end{array}$

$\text{Mole}\text{fraction}=\frac{\text{moles}\text{of}\text{solute}}{\text{total}\text{moles}}$

$\text{Moles}\text{of}\text{solute}\left({\text{NH}}_{\text{3}}\right)=\frac{252}{17}=14.82\text{moles}$$\left[\therefore \frac{\text{weight}}{\text{m}\text{.wt}}=\text{moles}\right]$

$\text{Moles}\text{of}\text{solvent}\left({\text{H}}_{\text{2}}\text{O}\right)=\frac{648}{18\text{g}}=36\text{moles}$

$\therefore \text{Mole}\text{fraction}\text{of}{\text{NH}}_{\text{3}}=\frac{\text{14}\text{.82}}{\text{14}\text{.82+36}}=0.2754$

Conclusion

The molarity, molality and mole fraction of the each reagent is as follows:

Reagents	Molarity (M)	Molality (m)	Mole fraction
Hydrochloric acid	12.38	16.97	0.23
Nitric acid	15.77	37.03	0.4
Sulfuric acid	17.83	193.8	0.7772
Acetic acid	17.33	1650	0.967
Ammonia	14.82	23.42	0.2754