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Chapter 13: Problem 63

In what volume of water must \(1 \mathrm{mol}\) of a nonelectrolyte be dissolved if the solution is to have an osmotic pressure of 1 atm at 273 K? Which of the gas laws does this result resemble?

Short Answer

Expert verified
The solution requires a volume of 22.4073 L of water. The result resembles the Ideal Gas Law.

Step by step solution

01

Reminder of the Osmotic Pressure Formula

Osmotic pressure (\(π\)) can be calculated using the formula: \(π = nRT/V\), where n is the number of moles, R is the ideal gas constant (0.0821 L.atm/mol.K), T is the temperature in Kelvin, and V is the volume.
02

Substituting the Known Values

Substituting the known values into the equation: \(1 = 1*0.0821*273/V\). We need to solve this equation for V (volume).
03

Solve for V (Volume)

To solve for V, we can apply algebraic operations. Multiplying the variables on the right side of the equation gives: 1 = 22.4073/V. Then, cross-multiplying yields V = 22.4073 L.
04

Examine the Result and Gas Laws

Looking at the result, it can be seen that this resembles the ideal gas law equation \(PV = nRT\), where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is temperature. In the osmotic pressure equation, the osmotic pressure (π) stands as equivalent to the pressure (P) in the ideal gas law.

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Most popular questions from this chapter

A benzene-toluene solution with \(x_{\text {benz }}=0.300\) has a normal boiling point of \(98.6^{\circ} \mathrm{C}\). The vapor pressure of pure toluene at \(98.6^{\circ} \mathrm{C}\) is \(533 \mathrm{mm} \mathrm{Hg}\). What must be the vapor pressure of pure benzene at \(98.6^{\circ} \mathrm{C} ?\) (Assume ideal solution behavior.)

Which aqueous solution from the column on the right has the property listed on the left? Explain your choices. $$\begin{array}{ll}\hline \text { Property } & \text { Solution } \\\\\hline \text { 1. lowest electrical } & \text { a. } 0.10 \mathrm{m} \mathrm{KCl}(\mathrm{aq}) \\\\\text { conductivity } & \\\\\text { 2. } \text { lowest boiling } & \text { b. } 0.15 \mathrm{m}\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(\mathrm{aq}) \\\\\text { point } & \\\\\text { 3. highest vapor pressure } & \text { c. } 0.10 \mathrm{m} \mathrm{CH}_{3} \mathrm{COOH}(\mathrm{aq}) \\\\\text { of water at } 25^{\circ} \mathrm{C} & \\\\\text { 4. lowest freezing point } & \text { d. } 0.05 \mathrm{m} \mathrm{NaCl} \\\\\hline\end{array}$$

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Styrene, used in the manufacture of polystyrene plastics, is made by the extraction of hydrogen atoms from ethylbenzene. The product obtained contains about \(38 \%\) styrene \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}=\mathrm{CH}_{2}\right)\) and \(62 \%\) ethylbenzene \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{CH}_{3}\right),\) by mass. The mixture is separated by fractional distillation at \(90^{\circ} \mathrm{C} .\) Determine the composition of the vapor in equilibrium with this \(38 \%-62 \%\) mixture at \(90^{\circ} \mathrm{C}\). The vapor pressure of ethylbenzene is \(182 \mathrm{mmHg}\) and that of styrene is \(134 \mathrm{mmHg}\).

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