Chapter 10, Problem 31E

Common commercial acids and bases are aqueous solutions with the following properties:

	Density (g/cm3)	Mass Percent of Solute
Hydrochloric acid	1.19	38
Nitric acid	1.42	70.
Sulfuric acid	1.84	95
Acetic acid	1.05	99
Ammonia	0.95	28

Calculate the molarity, molality, and mole fraction of each of the preceding reagents.

Interpretation Introduction

Interpretation: The molality, molarity and mole fraction of the reagents has to be calculated.

Concept Introduction:

Mole fraction of a compound can be defined as the number of moles of a substance to the total number of moles present in them. The mole fraction can be calculated by,

$\text{Mole}\text{fraction}\text{of}\text{compound=}\frac{\text{Number}\text{of}\text{moles(in}\text{mol)}}{\text{Total}\text{number}\text{of}\text{moles(in}\text{mol)}}$

Molality of a solution also called as molal concentration can be defined as the mass of solute in grams to the mass of the solvent in kilograms. It can be given by the equation,

$\text{Molality(mol}{\text{kg}}^{\text{-1}}\text{)=}\frac{\text{Mass}\text{of}\text{solue(in}\text{g)}}{\text{Mass}\text{of}\text{solvent(in}\text{kg)}}$

Molarity can be defined as the moles of solute (in grams) to the volume of the solution (in litres). The molarity of a solution can be given by the formula,

$\text{Molarity(M)=}\frac{\text{Moles}\text{of}\text{solute(in}\text{g)}}{\text{Volume}\text{of}\text{solution(in}\text{L)}}$

Answer to Problem 31E

Answer

Hydrochloric acid

    Molarity = $\text{12}\text{M}$

    Molality = $\text{17mol}{\text{kg}}^{\text{-1}}$

    Mole fraction = $\text{0}\text{.23}$

Nitric acid

   Molarity =$\text{16}\text{M}$

   Molality = $\text{37}\text{mol}{\text{kg}}^{\text{-1}}$

   Mole fraction = $\text{0}\text{.39}$

Sulphuric acid

   Molarity = $\text{18}\text{M}$

   Molality = $\text{200}\text{mol}{\text{kg}}^{\text{-1}}$

   Mole fraction = $0.76$

Acetic acid

   Molarity =$\text{17}\text{M}$

   Molality = $\text{2000}\text{mol}{\text{kg}}^{\text{-1}}$

   Mole fraction= $\text{0}\text{.96}$

Ammonia

   Molarity = $\text{15}\text{M}$

  Molality = $\text{23}\text{mol}{\text{kg}}^{\text{-1}}$

  Mole fraction = $\text{0}\text{.29}$

Explanation of Solution

Hydrochloric acid

Record the given data

Density of Hydrochloric acid = $\text{1}\text{.19}\text{g}{\text{cm}}^{\text{-3}}$

Mass percent of solute = $\text{38\%}$

To calculate the molarity of Hydrochloric acid

Molar mass of Hydrochloric acid = $\text{36}\text{.5}\text{g}$

Molarity of Hydrochloric acid =  $\frac{\text{38}\text{g}\text{HCl}}{\text{100}\text{.0}\text{g}}\text{×}\frac{\text{1}\text{.19}\text{g}\text{soln}}{{\text{cm}}^{\text{3}}\text{soln}}\text{×}\frac{\text{1000}{\text{cm}}^{\text{3}}}{\text{L}}\text{×}\frac{\text{1}\text{mol}\text{HCl}}{\text{36}\text{.5}\text{g}}$

                                            = $\text{12}\text{M}$

Molarity of Hydrochloric acid =$\text{12}\text{M}$

To calculate the molality of Hydrochloric acid

Molality of Hydrochloric acid = $\frac{\text{38}\text{g}\text{HCl}}{\text{62}\text{g}\text{solvent}}\text{×}\frac{\text{1000}\text{g}}{\text{kg}}\text{×}\frac{\text{1}\text{mol}\text{HCl}}{\text{36}\text{.5}\text{g}}$

                                             = $\text{17}\text{mol}{\text{kg}}^{\text{-1}}$

Molality of Hydrochloric acid = $\text{17}\text{mol}{\text{kg}}^{\text{-1}}$

To calculate the moles of Hydrochloric acid and Water

Molar mass of Hydrochloric acid = $\text{36}\text{.5}\text{g}$

Molar mass of Water = $\text{18}\text{.02}\text{g}$

Mass of solvent = $\text{62}\text{kg}$

Moles of Hydrochloric acid = $\text{38}\text{g}\text{HCl×}\frac{\text{1}\text{mol}}{\text{36}\text{.5}\text{g}}$

                                          = $\text{1}\text{.0}\text{mol}$

Moles of Water= $\text{62}\text{g}{\text{H}}_{\text{2}}\text{O×}\frac{\text{1}\text{mol}}{\text{18}\text{.0}\text{g}}$

                       = $\text{3}\text{.4}\text{mol}$

Moles of Hydrochloric acid = $\text{1}\text{.0}\text{mol}$

Moles of Water = $\text{3}\text{.4}\text{mol}$

To calculate the mole fraction of Hydrochloric acid

Moles of Hydrochloric acid = $\text{1}\text{.0}\text{mol}$

Moles of Water = $\text{3}\text{.4}\text{mol}$

Mole fraction of Hydrochloric acid = $\frac{\text{1}\text{.0}\text{mol}}{\text{4}\text{.4}\text{mol}}$

                                                     = $\text{0}\text{.23}$

Mole fraction of Hydrochloric acid=$\text{0}\text{.23}$

Nitric acid:

Record the given data

Density of Nitric acid = $\text{1}\text{.19}\text{g}{\text{cm}}^{\text{-3}}$

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

To calculate the molarity of Nitric acid

Molar mass of Nitric acid = $\text{63}\text{g}$

Molarity of Nitric acid = $\frac{\text{70}\text{.0}\text{g}{\text{HNO}}_{\text{3}}}{\text{100}\text{.}\text{g}\text{solution}}\text{×}\frac{\text{1}\text{.42}\text{g}\text{soln}}{{\text{cm}}^{\text{3}}\text{soln}}\text{×}\frac{\text{1000}{\text{cm}}^{\text{3}}}{\text{L}}\text{×}\frac{\text{1}\text{mol}{\text{HNO}}_{\text{3}}}{\text{63}\text{.0}\text{g}}$

                                  = $\text{16}\text{M}$

To calculate the molality of Nitric acid

Molality of Nitric acid = $\frac{\text{70g}{\text{HNO}}_{\text{3}}}{\text{30}\text{g}\text{solvent}}\text{×}\frac{\text{1000}\text{g}}{\text{kg}}\text{×}\frac{\text{1}\text{mol}{\text{HNO}}_{\text{3}}}{\text{63}\text{.0}\text{g}}$

                                             = $\text{37}\text{mol}{\text{kg}}^{\text{-1}}$

Molality of Nitric acid = $\text{37}\text{mol}{\text{kg}}^{\text{-1}}$

To calculate the moles of Nitric acid and Water

Molar mass of Nitric acid = $\text{63g}$

Molar mass of Water = $\text{18}\text{.02}\text{g}$

Mass of solvent = $\text{30kg}$

Moles of Nitric acid = $\text{70g}{\text{HNO}}_{\text{3}}\text{×}\frac{\text{1}\text{mol}}{\text{63}\text{.0g}}$

                                          = $\text{1}\text{.1mol}$

Moles of Water= $30\text{g}{\text{H}}_{\text{2}}\text{O×}\frac{\text{1}\text{mol}}{\text{18}\text{.0}\text{g}}$

                       = $1.7\text{mol}$

Moles of Nitric acid = $\text{1}\text{.1mol}$

Moles of Water = $1.7\text{mol}$

To calculate the mole fraction of Nitric acid

Moles of Nitric acid = $\text{1}\text{.1}\text{mol}$

Moles of Water = $1.7\text{mol}$

Mole fraction of Nitric acid = $\frac{\text{1}\text{.1mol}}{2.8\text{mol}}$

                                                     = $\text{0}\text{.39}$

Mole fraction of Nitric acid=$\text{0}\text{.39}$

Sulphuric acid:

Record the given data

Density of Sulphuric acid = $\text{1}\text{.84}\text{g}{\text{cm}}^{\text{-3}}$

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

To calculate the molarity of Sulphuric acid

Molar mass of Sulphuric acid = $\text{98}\text{.1}\text{g}$

Molarity of Sulphuric acid = $\frac{\text{95}\text{g}{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}}{\text{100}\text{g}\text{soln}}\text{×}\frac{\text{1}\text{.84}\text{g}\text{soln}}{{\text{cm}}^{\text{3}}\text{soln}}\text{×}\frac{\text{1000}{\text{cm}}^{\text{3}}}{\text{L}}\text{×}\frac{\text{1}\text{mol}{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}}{\text{98}\text{.1}\text{g}{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}}$

       =$\text{18}\text{M}$

Molarity of Sulphuric acid = $\text{18}\text{M}$

To calculate the molality of Sulphuric acid

Molality of Sulphuric acid = $\frac{\text{95g}{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}}{\text{5 g solvent}}\text{×}\frac{\text{1000}\text{g}}{\text{kg}}\text{×}\frac{\text{1}\text{mol}{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}}{\text{98}\text{.1}\text{g}}$

                                             = $\text{194}\text{mol}{\text{kg}}^{\text{-1}}\approx \text{200}\text{mol}\text{kg-1}$

Molality of Sulphuric acid = $\text{194}\text{mol}{\text{kg}}^{\text{-1}}\approx \text{200}\text{mol}\text{kg-1}$

To calculate the moles of Sulphuric acid and Water

Molar mass of Sulphuric acid = $\text{95}\text{g}$

Molar mass of Water = $\text{18}\text{.02}\text{g}$

Mass of solvent = $\text{5kg}$

Moles of Sulphuric acid = $\text{95g}{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\text{×}\frac{\text{1}\text{mol}}{\text{98}\text{.1g}}$

                                          = $\text{0}\text{.97mol}$

Moles of Water= $\text{5g}{\text{H}}_{\text{2}}\text{O×}\frac{\text{1}\text{mol}}{\text{18}\text{.0}\text{g}}$

                       = $\text{0}\text{.3}\text{mol}$

Moles of Sulphuric acid = $\text{0}\text{.97mol}$

Moles of Water = $\text{0}\text{.3}\text{mol}$

To calculate the mole fraction of Sulphuric acid

Moles of Sulphuric acid = $\text{0}\text{.97mol}$

Moles of Water = $\text{0}\text{.3}\text{mol}$

Mole fraction of Sulphuric acid = $\frac{\text{0}\text{.97mol}}{\text{1}\text{.27mol}}$

                                                 = $\text{0}\text{.76}$

Mole fraction of Sulphuric acid=$\text{0}\text{.76}$

Acetic acid:

Record the given data

Density of Acetic acid = $\text{1}\text{.05}\text{g}{\text{cm}}^{\text{-3}}$

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

To calculate the molarity of Acetic acid

Molar mass of Acetic acid = $\text{60}\text{.05}\text{g}$

Molarity of Acetic acid = $\frac{\text{99}\text{g}{\text{CH}}_{\text{3}}\text{COOH}}{\text{100}\text{g}\text{soln}}\text{×}\frac{\text{1}\text{.05}\text{g}\text{soln}}{{\text{cm}}^{\text{3}}\text{soln}}\text{×}\frac{\text{1000}{\text{cm}}^{\text{3}}}{\text{L}}\text{×}\frac{\text{1}\text{mol}{\text{CH}}_{\text{3}}\text{COOH}}{\text{60}\text{.05g}{\text{CH}}_{\text{3}}\text{COOH}}$

       =$\text{17}\text{M}$

Molarity of Acetic acid = $\text{17}\text{M}$

To calculate the molality of Acetic acid

Molality of Acetic acid = $\frac{\text{99g}{\text{CH}}_{\text{3}}\text{COOH}}{\text{1g solvent}}\text{×}\frac{\text{1000}\text{g}}{\text{kg}}\text{×}\frac{\text{1}\text{mol}{\text{CH}}_{\text{3}}\text{COOH}}{\text{60}\text{.05}\text{g}}$

                                             = $\text{1600}\text{mol}{\text{kg}}^{\text{-1}}\approx \text{2000}\text{mol}\text{kg-1}$

Molality of Acetic acid = $\text{1600}\text{mol}{\text{kg}}^{\text{-1}}\approx \text{2000}\text{mol}\text{kg-1}$

To calculate the moles of Acetic acid and Water

Molar mass of Acetic acid = $\text{60}\text{.05}\text{g}$

Molar mass of Water = $\text{18}\text{.02}\text{g}$

Mass of solvent = $\text{1kg}$

Moles of Acetic acid = $\text{99g}{\text{CH}}_{\text{3}}\text{COOH×}\frac{\text{1}\text{mol}}{\text{60}\text{.01g}}$

                                          = $\text{1}\text{.6}\text{mol}$

Moles of Water= $\text{1g}{\text{H}}_{\text{2}}\text{O×}\frac{\text{1}\text{mol}}{\text{18}\text{.0}\text{g}}$

                       = $\text{0}\text{.06}\text{mol}$

Moles of Acetic acid = $\text{1}\text{.6}\text{mol}$

Moles of Water = $\text{0}\text{.06}\text{mol}$

To calculate the mole fraction of Acetic acid

Moles of Acetic acid = $\text{1}\text{.6}\text{mol}$

Moles of Water = $\text{0}\text{.06}\text{mol}$

Mole fraction of Acetic acid = $\frac{\text{1}\text{.6 mol}}{\text{1}\text{.66mol}}$

                                                 = $\text{0}\text{.96}$

Mole fraction of Acetic acid=$\text{0}\text{.96}$

Ammonia:

Record the given data

Density of Ammonia= $\text{0}\text{.90}\text{g}{\text{cm}}^{\text{-3}}$

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

To calculate the molarity of Ammonia

Molar mass of Ammonia = $\text{17}\text{.0}\text{g}$

Molarity of Ammonia = $\frac{\text{28}\text{g}{\text{NH}}_{\text{3}}}{\text{100}\text{g}\text{soln}}\text{×}\frac{\text{0}\text{.90}\text{g}}{{\text{cm}}^{\text{3}}}\text{×}\frac{\text{1000}{\text{cm}}^{\text{3}}}{\text{L}}\text{×}\frac{\text{1}\text{mol}}{\text{17}\text{.0}\text{g}}$

                                  =$\text{15}\text{M}$

Molarity of Ammonia=$\text{15}\text{M}$

To calculate the molality of Ammonia

Molality of Ammonia= $\frac{\text{28g}{\text{NH}}_{\text{3}}}{\text{72g solvent}}\text{×}\frac{\text{1000}\text{g}}{\text{kg}}\text{×}\frac{\text{1}\text{mol}{\text{NH}}_{\text{3}}}{\text{17}\text{.00}\text{g}}$

                                 = $\text{23}\text{mol}{\text{kg}}^{\text{-1}}$

Molality of Ammonia = $\text{23}\text{mol}{\text{kg}}^{\text{-1}}$

To calculate the moles of Ammonia and Water

Molar mass of Ammonia= $\text{17}\text{.0g}$

Molar mass of Water = $\text{18}\text{.02}\text{g}$

Mass of solvent = $\text{72kg}$

Moles of Ammonia= $\text{28g}{\text{NH}}_{\text{3}}\text{×}\frac{\text{1}\text{mol}}{\text{17}\text{.0g}}$

                                          = $\text{1}\text{.6}\text{mol}$

Moles of Water= $\text{72g}{\text{H}}_{\text{2}}\text{O×}\frac{\text{1}\text{mol}}{\text{18}\text{.0}\text{g}}$

                       = $4.00\text{mol}$

Moles of Ammonia = $\text{1}\text{.6}\text{mol}$

Moles of Water = $4.00\text{mol}$

To calculate the mole fraction of Ammonia

Moles of Ammonia= $\text{1}\text{.6}\text{mol}$

Moles of Water = $\text{4}\text{.00mol}$

Mole fraction of Ammonia = $\frac{\text{1}\text{.6 mol}}{\text{5}\text{.60mol}}$

                                         = $\text{0}\text{.29}$

Mole fraction of Ammonia=$\text{0}\text{.29}$

Conclusion

The molality, molarity and mole fraction of the given reagents were calculated.