Chapter 13, Problem 56QRT

A 12-oz (355-mL) Pepsi contains 38.9 mg caffeine (molar mass = 194.2 g/mol). Assume that the Pepsi, mainly water, has a density of 1.01 g/mL. For such a Pepsi, calculate: (a) its caffeine concentration in ppm; (b) its molarity of caffeine; and (c) the molality of caffeine.

Interpretation Introduction

Interpretation:

The caffeine concentration in $\text{ppm}$ has to be calculated.

Answer to Problem 56QRT

The caffeine concentration in $\text{ppm}$ was $108ppm.$

Explanation of Solution

Given data as follows: Molar mass is $\text{194}\text{.2}\text{g/mol}$; density of water is $\text{1}\text{.0}\text{g/mL}$; mass of caffeine is $\text{38}\text{.9}\text{mg}\text{.}$

As known, $\text{1 ppm = 1 mg/1kg}\text{.}$

Conversion of water from $\text{ml into kg}$ as,

  $\begin{array}{l}\text{Volume =}\text{355 mL}\\ \\ \text{= 355 mL}\text{solution}\text{×}\frac{\text{1}\text{.01}\text{g}\text{solution}}{\text{1}\text{mL}\text{solution}}\text{×}\frac{\text{1}\text{kg}}{\text{1000}\text{g}}\text{= 0}\text{.359}\text{kg}\text{solution}\text{.}\end{array}$

Conversion of $\text{mole into ppm}$ into $\text{ppm}$ as,

  $\begin{array}{l}\text{1 ppm = 1 mg/1kg}\text{.}\\ \\ \text{=}\frac{\text{38}\text{.9}\text{×}{\text{10}}^{\text{-3}}\text{g}}{\text{0}\text{.359}\text{kg}\text{water}}\text{=}108ppm.\end{array}$

Therefore, caffeine concentration in $\text{ppm}$ was $108ppm.$

Interpretation Introduction

Interpretation:

The molarity of caffeine has to be calculated.

Concept introduction:

Molarity: Dividing moles of solute by the volume of solution in Liter.

  $\text{Molarity}\text{=}\frac{\text{Moles}\text{of}\text{solute}}{\text{Volume of solution in L}}.$

Answer to Problem 56QRT

The caffeine concentration in $\text{ppm}$ was $108ppm.$

Explanation of Solution

Given data as follows: Molar mass is $\text{194}\text{.2}\text{g/mol}$; density of water is $\text{1}\text{.0}\text{g/mL}$; mass of caffeine is $\text{38}\text{.9}\text{mg}$; volume of Pepsi is $\text{355}\text{mL}\text{.}$

Conversion of mass into mole as,

  $\begin{array}{c}\text{Moles}\text{=}\frac{\text{Mass}\text{of}\text{caffeine}}{\text{Molar}\text{Mass}\text{of}\text{caffeine}}\\ \\ \text{=}\frac{\text{38}\text{.9}\text{×}{\text{10}}^{\text{-3}}\text{g}}{\text{194}\text{.2}\text{g}\text{/mol}}\text{=}\text{2}\text{.003}\text{×}{\text{10}}^{\text{-4}}\text{mol}\text{.}\end{array}$

Conversion of $\text{ml}$ into liter as,

  $\text{355}\text{mL}\text{solution×}\frac{\text{1L}}{\text{1000}\text{mL}}\text{=}\text{0}\text{.355}\text{L}\text{.}$

Molarity of solution is calculated as,

  $\text{Molarity}\text{=}\frac{\text{2}\text{.003}\text{×}{\text{10}}^{\text{-4}}\text{mol}}{\text{0}\text{.355}\text{L}}\text{=}5.64\times 10^{-4}M.$

Therefore, molarity of caffeine was $5.64\times 10^{-4}M.$

Interpretation Introduction

Interpretation:

The molality of caffeine has to be calculated.

Concept introduction:

Molality: Dividing the moles of solute by the mass of solvent in kg.

  $\text{Molality}\text{=}\frac{\text{Moles}\text{of}\text{solute}}{\text{Mass}\text{of}\text{solvent}\text{in kg}}.$

Answer to Problem 56QRT

The molality of caffeine was $5.58\times 10^{-4}m.$

Explanation of Solution

Given data as follows: Molar mass is $\text{194}\text{.2}\text{g/mol}$; density of water is $\text{1}\text{.0}\text{g/mL}$; mass of caffeine is $\text{38}\text{.9}\text{mg}$; volume of Pepsi is $\text{355}\text{mL}\text{.}$

Conversion of mass into mole as,

  $\begin{array}{c}\text{Moles}\text{=}\frac{\text{Mass}\text{of}\text{caffeine}}{\text{Molar}\text{Mass}\text{of}\text{caffeine}}\\ \\ \text{=}\frac{\text{38}\text{.9}\text{×}{\text{10}}^{\text{-3}}\text{g}}{\text{194}\text{.2}\text{g}\text{/mol}}\text{=}\text{2}\text{.003}\text{×}{\text{10}}^{\text{-4}}\text{mol}\text{.}\end{array}$

Conversion of $\text{ml}$ into kg as,

Conversion of water from $\text{ml}$ into kg as,

  $\begin{array}{c}\text{Volume =}\text{355 mL}\\ \\ \text{= 355 mL}\text{solution}\text{×}\frac{\text{1}\text{.01}\text{g}\text{solution}}{\text{1}\text{mL}\text{solution}}\text{×}\frac{\text{1}\text{kg}}{\text{1000}\text{g}}\text{= 0}\text{.359}\text{kg}\text{solution}\text{.}\end{array}$

Molality of solution is calculated as,

$\text{Molality}\text{=}\frac{\text{2}\text{.003}\text{×}{\text{10}}^{\text{-4}}\text{mol}}{\text{0}\text{.359}\text{kg}}\text{=}5.58\times 10^{-4}m.$

Therefore, molality of caffeine was $5.58\times 10^{-4}m.$