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Chapter 11: Problem 133

A 1.60 -g sample of a mixture of naphthalene $\left(\mathrm{C}_{10} \mathrm{H}_{8}\right)\( and anthracene \)\left(\mathrm{C}_{14} \mathrm{H}_{10}\right)\( is dissolved in 20.0 \)\mathrm{g}$ benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) The freezing point of the solution is \(2.81^{\circ} \mathrm{C} .\) What is the composition as mass percent of the sample mixture? The freezing point of benzene is $5.51^{\circ} \mathrm{C}\( and \)K_{\mathrm{f}}\( is \)5.12^{\circ} \mathrm{C} \cdot \mathrm{kg} / \mathrm{mol} .$

Short Answer

Expert verified
The composition of the sample mixture is approximately \(52.87 \%\) naphthalene and \(47.13 \%\) anthracene.

Step by step solution

01

Calculate the freezing point depression

To calculate the freezing point depression, we subtract the freezing point of the pure solvent from the freezing point of the solution and use the given values in the exercise: Freezing point depression, \(\Delta T_f = T_{f_{solvent}} - T_{f_{solution}}\) \(\Delta T_f = 5.51 ^\circ \mathrm{C} - 2.81 ^\circ \mathrm{C} = 2.7 ^\circ \mathrm{C}\)
02

Calculate the molality of the solution

Using the freezing point depression, we can now find the molality (moles of solute per kilogram of solvent) of the solution using the formula: \(molality = \dfrac{\Delta T_f}{K_f}\) \(molality = \dfrac{2.7 ^\circ \mathrm{C}}{5.12 ^\circ \mathrm{C} \cdot \mathrm{kg} / \mathrm{mol}} = 0.527 \space \mathrm{mol/kg}\)
03

Set up equations to solve for mass of naphthalene and anthracene

Let \(x\) be the mass of naphthalene and \(y\) be the mass of anthracene present in the mixture. We can now create two equations using the information we know about the mass and molality of the solution: 1) Total mass of the mixture: \(x + y = 1.60 \space \mathrm{g}\) 2) Total moles of solute: \(\dfrac{x}{M_{naphthalene}} + \dfrac{y}{M_{anthracene}} = molality \times \frac{mass_{solvent}}{1000}\) Here, \(M_{naphthalene}\) is the molar mass of naphthalene (\(128.17 \space \mathrm{g/mol}\)) and \(M_{anthracene}\) is the molar mass of anthracene (\(178.23 \space \mathrm{g/mol}\)).
04

Solve the equations

Now we can solve these equations to find the mass of naphthalene and anthracene: From equation 1, we can rewrite as: \(y = 1.60 - x\) Substitute this value of \(y\) into equation 2: \(\dfrac{x}{128.17} + \dfrac{1.60 - x}{178.23} = 0.527 \times \dfrac{20}{1000}\) Now we can solve the equation for \(x\): \(x = 0.846 \space \mathrm{g}\) Using this value of \(x\) and equation 1, we can find the value of \(y\): \(y = 1.60 - 0.846\) \(y = 0.754 \space \mathrm{g}\)
05

Calculate the mass percent of naphthalene and anthracene

To find the mass percent, we use the formula: Mass percent \(= \dfrac{mass \space of \space solute}{total \space mass \space in \space mixture} \times 100\) For naphthalene: Mass percent \(= \dfrac{0.846}{1.60} \times 100 = 52.87 \% \) For anthracene: Mass percent \(= \dfrac{0.754}{1.60} \times 100 = 47.13 \% \) Therefore, the composition of the sample mixture is approximately \(52.87 \%\) naphthalene and \(47.13 \%\) anthracene.

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Most popular questions from this chapter

Plants that thrive in salt water must have internal solutions (inside the plant cells) that are isotonic with (have the same osmotic pressure as) the surrounding solution. A leaf of a saltwater plant is able to thrive in an aqueous salt solution (at \(25^{\circ} \mathrm{C} )\) that has a freezing point equal to \(-0.621^{\circ} \mathrm{C} .\) You would like to use this information to calculate the osmotic pressure of the solution in the cell. a. In order to use the freezing-point depression to calculate osmotic pressure, what assumption must you make (in addition to ideal behavior of the solutions, which we will assume)? b. Under what conditions is the assumption (in part a) reasonable? c. Solve for the osmotic pressure (at \(25^{\circ} \mathrm{C} )\) of the solution in the plant cell. d. The plant leaf is placed in an aqueous salt solution (at $25^{\circ} \mathrm{C}\( ) that has a boiling point of \)102.0^{\circ} \mathrm{C} .$ What will happen to the plant cells in the leaf?

You and your friend are each drinking cola from separate 2-L bottles. Both colas are equally carbonated. You are able to drink 1 L of cola, but your friend can drink only about half a liter. You each close the bottles and place them in the refrigerator. The next day when you each go to get the colas, whose will be more carbonated and why?

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The normal boiling point of diethyl ether is \(34.5^{\circ} \mathrm{C}\) . A solution containing a nonvolatile solute dissolved in diethyl ether has a vapor pressure of 698 torr at \(34.5^{\circ} \mathrm{C} .\) What is the mole fraction of diethyl ether in this solution?
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