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Chapter 13: Problem 48

Ascorbic acid (vitamin C, \(\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{6}\) ) is a water-soluble vitamin. A solution containing \(80.5 \mathrm{~g}\) of ascorbic acid dissolved in \(210 \mathrm{~g}\) of water has a density of \(1.22 \mathrm{~g} / \mathrm{mL}\) at \(55^{\circ} \mathrm{C}\). Calculate (a) the mass percentage, \((\mathbf{b})\) the mole fraction, \((\mathbf{c})\) the molality, \((\mathbf{d})\) the molarity of ascorbic acid in this solution.

Short Answer

Expert verified
a) 27.7%, b) 0.0378, c) 2.18 mol/kg, d) 1.92 M.

Step by step solution

01

Calculate the Mass Percentage

The mass percentage is given by the formula:\[\text{Mass Percentage} = \left( \frac{\text{Mass of Solute}}{\text{Total Mass of Solution}} \right) \times 100\%\]Substitute the given values:\[= \left( \frac{80.5 \text{ g}}{80.5 \text{ g} + 210 \text{ g}} \right) \times 100\%\]\[= \left( \frac{80.5}{290.5} \right) \times 100\% \approx 27.7\% \]
02

Calculate the Mole Fraction

First, calculate the moles of each component. The molar mass of ascorbic acid (\(\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{6}\)) is approximately 176.12 g/mol.Moles of ascorbic acid:\[= \frac{80.5 \text{ g}}{176.12 \text{ g/mol}} \approx 0.457 \text{ mol}\]The molar mass of water \((\mathrm{H}_{2}\mathrm{O})\) is 18.02 g/mol, so moles of water:\[= \frac{210 \text{ g}}{18.02 \text{ g/mol}} \approx 11.66 \text{ mol}\]Now calculate the mole fraction of ascorbic acid:\[\text{Mole Fraction of Ascorbic Acid} = \frac{0.457}{0.457 + 11.66} \approx 0.0378\]
03

Calculate the Molality

Molality is calculated using the formula:\[\text{Molality (m)} = \frac{\text{Moles of solute}}{\text{Mass of solvent in kg}}\]Substitute the given values:\[= \frac{0.457}{0.210} \approx 2.18 \text{ mol/kg}\]
04

Calculate the Molarity

First, find the volume of the solution using its density. The total mass of the solution is 290.5 g, and its density is 1.22 g/mL:\[\text{Volume of solution} = \frac{290.5 \text{ g}}{1.22 \text{ g/mL}} = 238.11 \text{ mL} = 0.23811 \text{ L}\]Molarity is calculated using:\[\text{Molarity (M)} = \frac{\text{Moles of solute}}{\text{Volume of solution in L}}\]Substitute the values:\[= \frac{0.457}{0.23811} \approx 1.92 \text{ M}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Mass Percentage
When thinking about mass percentage, it helps to picture it as the slice of the total pie represented by the ingredient you're interested in. In this case, we want to know how much of the total solution is made up by ascorbic acid.
We'll use the formula:
  • Mass Percentage = \( \left( \frac{\text{Mass of Solute}}{\text{Total Mass of Solution}} \right) \times 100\% \)
You simply take the mass of ascorbic acid (our solute) over the total mass of the solution (ascorbic acid plus water), then multiply by 100 to get a percentage. This gives you the percentage of the solution that is ascorbic acid.
Here, it turns out to be about 27.7%, meaning out of 100 parts of the solution, roughly 27.7 parts are ascorbic acid. It's a useful way to express concentration when you're dealing with solutions.
Mole Fraction
Mole fraction is a way to express concentration by comparing the number of moles of one component to the total number of moles in the solution. It's all about the ratio of particles rather than mass.
  • Mole Fraction = \( \frac{\text{Moles of Solute}}{\text{Total Moles in Solution}} \)
For this exercise, we calculate the moles of both ascorbic acid and water separately, since they are both contributing to the total number of moles in the solution. We found that ascorbic acid accounts for about 0.0378 of the mole fraction.
This means that approximately 3.78% of all the particles in this solution are ascorbic acid molecules. Unlike mass percentage, mole fraction doesn't deal with weights, it focuses purely on the count of molecules.
Molality
Molality provides another angle on concentration, centered around the number of moles of solute per kilogram of solvent.
It is distinct from molarity, as molality depends on the mass of the solvent, and not the total solution volume. Here鈥檚 the formula:
  • Molality (m) = \( \frac{\text{Moles of solute}}{\text{Mass of solvent in kg}} \)
This measure comes in handy when temperature and pressure fluctuations occur, because unlike volume, mass remains stable under such changes. For this solution, the molality is about 2.18 mol/kg. This tells us that there are 2.18 moles of ascorbic acid for every kilogram of water acting as the solvent.
This measure is crucial in situations where you need precision regardless of environmental conditions.
Molarity
Among the more common ways to express solution concentration, molarity focuses on the number of moles of solute within a specific volume of solution. It's used in many chemical applications and is relatively easy to calculate.
The formula is:
  • Molarity (M) = \( \frac{\text{Moles of solute}}{\text{Volume of solution in L}} \)
For our ascorbic acid solution, once we calculated the volume from the given total mass and density, we found the molarity to be about 1.92 M. This means there are 1.92 moles of ascorbic acid in every liter of solution.
Molarity is commonly used in laboratory settings due to its direct relationship with reaction stoichiometry, via volume-based dosing.

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Most popular questions from this chapter

\(\mathrm{KBr}\) is relatively soluble in water, yet its enthalpy of solution is \(+19.8 \mathrm{~kJ} / \mathrm{mol}\). Which of the following statements provides the best explanation for this behavior? (a) Potassium salts are always soluble in water. (b) The entropy of mixing must be unfavorable. (c) The enthalpy of mixing must be small compared to the enthalpies for breaking up water-water interactions and K-Br ionic interactions. (d) \(\mathrm{KBr}\) has a high molar mass compared to other salts like \(\mathrm{NaCl}\)

When ammonium chloride dissolves in water, the solution becomes colder. (a) Is the solution process exothermic or endothermic? (b) Why does the solution form?

An ionic compound has a very negative \(\Delta H_{\text {soln }}\) in water. (a) Would you expect it to be very soluble or nearly insoluble in water? (b) Which term would you expect to be the largest negative number: \(\Delta H_{\text {solvent }}, \Delta H_{\text {solute }}\), or \(\Delta H_{\operatorname{mix}} ?\)

The Henry's law constant for hydrogen gas \(\left(\mathrm{H}_{2}\right)\) in water at \(25^{\circ} \mathrm{C}\) is \(7.7 \times 10^{-6} \mathrm{M} / \mathrm{kPa}\) and the constant for argon \((\mathrm{Ar})\) at \(25^{\circ} \mathrm{C}\) is \(1.4 \times 10^{-5} \mathrm{M} / \mathrm{kPa}\). If the two gases are each present at 253 kPa pressure, calculate the solubility of each gas.

The following table presents the solubilities of several gases in water at \(25^{\circ} \mathrm{C}\) under a total pressure of gas and water vapor of \(101.3 \mathrm{kPa}\). (a) What volume of \(\mathrm{CH}_{4}(g)\) under standard conditions of temperature and pressure is contained in \(4.0 \mathrm{~L}\) of a saturated solution at \(25^{\circ} \mathrm{C} ?\) (b) The solubilities (in water) of the hydrocarbons are as follows: methane \(<\) ethane \(<\) ethylene. Is this because ethylene is the most polar molecule? (c) What intermolecular interactions can these hydrocarbons have with water? (d) Draw the Lewis dot structures for the three hydrocarbons. Which of these hydrocarbons possess \(\pi\) bonds? Based on their solubilities, would you say \(\pi\) bonds are more or less polarizable than \(\sigma\) bonds? (e) Explain why \(\mathrm{NO}\) is more soluble in water than either \(\mathrm{N}_{2}\) or \(\mathrm{O}_{2}\). (f) \(\mathrm{H}_{2} \mathrm{~S}\) is more water-soluble than almost all the other gases in table. What intermolecular forces is \(\mathrm{H}_{2} \mathrm{~S}\) likely to have with water? \((\mathbf{g}) \mathrm{SO}_{2}\) is by far the most water-soluble gas in table. What intermolecular forces is \(\mathrm{SO}_{2}\) likely to have with water? $$ \begin{array}{lc} \hline \text { Gas } & \text { Solubility (mM) } \\ \hline \mathrm{CH}_{4} \text { (methane) } & 1.3 \\ \mathrm{C}_{2} \mathrm{H}_{6} \text { (ethane) } & 1.8 \\ \mathrm{C}_{2} \mathrm{H}_{4} \text { (ethylene) } & 4.7 \\ \mathrm{~N}_{2} & 0.6 \\ \mathrm{O}_{2} & 1.2 \\ \mathrm{NO} & 1.9 \\ \mathrm{H}_{2} \mathrm{~S} & 99 \\ \mathrm{SO}_{2} & 1476 \\ \hline \end{array} $$
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