Chapter 17, Problem 21E

Interpretation Introduction

Interpretation: The density, mole fraction, molarity and molality of the given solution needs to be determined.

Concept introduction: The density, of a substance is defined as mass per unit volume and is given as,

  $\text{ρ=}\frac{\text{m}}{\text{v}}$

Mole fraction is given as,

  $Molefractionofsolute=\frac{MolesofSolute}{\begin{array}{l}MolesofSolute+MolesofSolvent\\ =\frac{n_A}{n_A+n_B}\end{array}}$

Molarity is given as:

  $M=\frac{n}{v}$

Here,

  $M$ = Molar concentration

  $n$ = moles of solute

  $v$ = liters of solution

Molality is given as:

  $molality(m)=\frac{numberofmolesofsolute(n)}{weightofthesolventinKg}$

Answer to Problem 21E

Density of the given solution is = $\text{1}\text{.057}\frac{\text{g}}{\text{mL}}$

Mole fraction of the solute = 0.180

Mole fraction of the solvent = 0.98

Molarity of the solution = 0.98 mol/L

Molality of the solution =1.20 mol/kg

Explanation of Solution

Given,

Mass of phosphoric acid=10.0 g

Volume of water = 100 mL

Total volume of solution = 104.0 mL

Density of water = $\text{1}\text{.00}\frac{\text{g}}{{\text{cm}}^{\text{3}}}$

Mass corresponding to 100.0 mL of water = $\text{Density of water×}\text{volume}\text{of}\text{water}$

  $\begin{array}{l}=1.00\frac{\text{g}}{{\text{cm}}^{\text{3}}}\times 100\\ =1.00\text{g}\left(\text{Since}{\text{cm}}^{\text{3}}\text{=}\text{mL}\right)\end{array}$

  $\begin{array}{l}\therefore \text{Total}\text{mass}\text{of}\text{solution}\text{=}\text{1}\text{.00}\text{g}\text{+}\text{10}\text{.0}\text{g}\\ \text{=}\text{110}\text{.0}\text{g}\end{array}$

Total volume of solution = 104.0 mL

  $\begin{array}{l}\therefore \text{Density}\text{=}\frac{\text{110}\text{g}}{\text{104}\text{mL}}\\ \text{=}\text{1}\text{.057}\frac{\text{g}}{\text{mL}}\end{array}$

  $\begin{array}{l}\text{Number}\text{of}\text{moles}\text{of}\text{solute}\text{=}\frac{\text{Mass}\text{of}{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}\text{in}\text{g}}{\text{Molar}\text{mass}\text{of}{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}\text{in}\text{g/mol}}\\ \text{=}\frac{\text{10}\text{g}}{\text{97}\text{.994}\text{g/mol}}\\ \text{=}\text{0}\text{.102}\text{moles}\end{array}$

  $\begin{array}{l}\text{Number}\text{of}\text{moles}\text{of}\text{solvent}\text{=}\frac{\text{Mass}\text{of}{\text{H}}_2\text{O}\text{in}\text{g}}{\text{Molar}\text{mass}\text{of}{\text{H}}_2{\text{O}}_{}\text{in}\text{g/mol}}\\ \text{=}\frac{\text{100}\text{g}}{18.015\text{g/mol}}\\ \text{=}5.55\text{moles}\end{array}$

  $\begin{array}{l}\therefore \text{Total}\text{moles}\text{=}\text{0}\text{.102}\text{moles}\text{+}\text{5}\text{.55}\text{moles}\\ \text{=}\text{5}\text{.652}\text{moles}\end{array}$

  $\begin{array}{l}\text{Mole}\text{fraction}\text{of}\text{solute}\text{=}\frac{\text{Number}\text{of}\text{moles}\text{of}\text{solute}}{\text{Total}\text{moles}}\\ \text{=}\frac{\text{0}\text{.102}\text{moles}}{\text{5}\text{.65}\text{moles}}\\ =0.0180\end{array}$

  $\begin{array}{l}\text{Mole}\text{fraction}\text{of}\text{solvent}\text{=}\frac{\text{Number}\text{of}\text{moles}\text{of}\text{solvent}}{\text{Total}\text{moles}}\\ \text{=}\frac{\text{5}\text{.55}\text{moles}}{\text{5}\text{.65}\text{moles}}\\ \text{=}\text{0}\text{.982}\end{array}$

  $\begin{array}{l}\text{Molarity}\text{=}\frac{\text{Number}\text{of}\text{moles}\text{of}\text{solute}}{\text{Volume}\text{of}\text{solution}\text{in}\text{L}}\\ \text{=}\frac{\text{0}\text{.102}\text{moles}}{\text{0}\text{.104}\text{L}}\\ \text{=}\text{0}\text{.98}\text{mol/L}\end{array}$

  $\begin{array}{l}\text{Molality}\text{=}\frac{\text{Number}\text{of}\text{moles}\text{of}\text{solute}}{\text{Mass}\text{of}\text{solvent}\text{in}\text{Kg}}\\ \text{=}\frac{\text{0}\text{.102}\text{moles}}{\text{0}\text{.100}\text{kg}}\\ \text{=}\text{1}\text{.20}\text{mol/kg}\end{array}$

Conclusion

Density of the given solution is = $\text{1}\text{.057}\frac{\text{g}}{\text{mL}}$

Mole fraction of the solute = 0.180

Mole fraction of the solvent = 0.98

Molarity of the solution = 0.98 mol/L

Molality of the solution =1.20 mol/kg