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Chapter 11: Problem 62

At a certain temperature, the vapor pressure of pure benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) is 0.930 \(\mathrm{atm} .\) A solution was prepared by dissolving 10.0 \(\mathrm{g}\) of a nondissociating, nonvolatile solute in 78.11 \(\mathrm{g}\) of benzene at that temperature. The vapor pressure of the solution was found to be 0.900 \(\mathrm{atm}\) . Assuming the solution behaves ideally, determine the molar mass of the solute.

Short Answer

Expert verified
The molar mass of the solute is approximately 299 g/mol.

Step by step solution

01

Raoult's law for the solution

According to Raoult's law, the vapor pressure of the solution (P_solution) can be written as the product of the mole fraction of the solvent (X_benzene) and the vapor pressure of the pure solvent (P_pure_benzene): \[P_{solution} = X_{benzene} \cdot P_{pure\_benzene}\]
02

Calculate the mole fraction of benzene in the solution

We are given the vapor pressure of the pure benzene (P_pure_benzene) and the vapor pressure of the solution (P_solution). We can rearrange Raoult's law to find the mole fraction of benzene (X_benzene): \[X_{benzene} = \frac{P_{solution}}{P_{pure\_benzene}}\] Now, plug in the given values to find X_benzene: \[X_{benzene} = \frac{0.900 atm}{0.930 atm} = 0.9677\]
03

Calculate the mole fraction of the solute (X_solute)

Since there are only two components in the solution (benzene and the solute), their mole fractions must add up to 1: \[X_{solute} = 1 - X_{benzene}\] Using the value of X_benzene calculated in step 2: \[X_{solute} = 1 - 0.9677 = 0.0323\]
04

Calculate the moles of benzene and the solute in the solution

We are given the mass of benzene (78.11 g) and can look up its molar mass (78.114 g/mol): moles of benzene = mass_benzene / molar_mass_benzene moles of benzene = 78.11 g / 78.114 g/mol = 1 mol Next, we can find the moles of the solute: moles of solute = moles of benzene * (X_solute / X_benzene) moles of solute = 1 mol * (0.0323 / 0.9677) = 0.0334 mol
05

Calculate the molar mass of the solute

We are given the mass of the solute (10.0 g). Using the moles of the solute calculated in step 4, find the molar mass of the solute: molar_mass_solute = mass_solute / moles of solute molar_mass_solute = 10.0 g / 0.0334 mol = 299 g/mol The molar mass of the solute is approximately 299 g/mol.

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Most popular questions from this chapter

For each of the following pairs, predict which substance is more soluble in water. a. \(\mathrm{CH}_{3} \mathrm{NH}_{2}\) or \(\mathrm{NH}_{3}\) b. \(\mathrm{CH}_{3} \mathrm{CN}\) or \(\mathrm{CH}_{3} \mathrm{OCH}_{3}\) c. \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\) or $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{3}$ d. \(\mathrm{CH}_{3} \mathrm{OH}\) or $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}$ e. \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCH}_{2} \mathrm{OH}\) or \(\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{6} \mathrm{OH}\) f. \(\mathrm{CH}_{3} \mathrm{OCH}_{3}\) or $\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}$

Calculate the solubility of \(\mathrm{O}_{2}\) in water at a partial pressure of \(\mathrm{O}_{2}\) of 120 torr at \(25^{\circ} \mathrm{C}\) . The Henry's law constant for \(\mathrm{O}_{2}\) is $1.3 \times 10^{-3} \mathrm{mol} / \mathrm{L} \cdot\( atm for Henry's law in the form \)C=k P\( where \)C$ is the gas concentration \((\mathrm{mol} / \mathrm{L})\)

What volume of ethylene glycol $\left(\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}_{2}\right),$ a nonelectrolyte,must be added to 15.0 \(\mathrm{L}\) water to produce an antifreeze solution with a freezing point of \(-25.0^{\circ} \mathrm{C} ?\) What is the boiling point of this solution? (The density of ethylene glycol is \(1.11 \mathrm{g} / \mathrm{cm}^{3},\) and the density of water is 1.00 \(\mathrm{g} / \mathrm{cm}^{3} . )\)

You make \(20.0 \mathrm{~g}\) of a sucrose $\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\( and \)\mathrm{NaCl}$ mixture and dissolve it in \(1.00 \mathrm{~kg}\) water. The freezing point of this solution is found to be \(-0.426^{\circ} \mathrm{C}\). Assuming ideal behavior, calculate the mass percent composition of the original mixture, and the mole fraction of sucrose in the original mixture.

An aqueous antifreeze solution is 40.0\(\%\) ethylene glycol \(\left(\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}_{2}\right)\) by mass. The density of the solution is 1.05 \(\mathrm{g} / \mathrm{cm}^{3}\) . Calculate the molality, molarity, and mole fraction of the ethylene glycol.
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