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The molar mass of a polymer was determined by measuring the osmotic pressure, \(7.6 \mathrm{mmHg},\) of a benzene solution containing \(5.0 \mathrm{~g}\) of the polymer dissolved in \(1.0 \mathrm{~L}\) solution. Calculate the molar mass of the polymer. Assume a temperature of \(298.15 \mathrm{~K}\).

Short Answer

Expert verified
The molar mass of the polymer is approximately 12,255 g/mol.

Step by step solution

01

Understand the Van't Hoff Equation for Osmotic Pressure

The Van't Hoff equation for osmotic pressure (\(\pi\)) is given by:\[\pi = \frac{n}{V}RT\]where \(\pi\) is the osmotic pressure, \(n\) is the number of moles of solute, \(V\) is the volume of the solution in liters, \(R\) is the universal gas constant \((0.0821\, \text{L atm K}^{-1} \text{mol}^{-1})\), and \(T\) is the temperature in Kelvin.
02

Convert Osmotic Pressure to Atmospheres

Osmotic pressure is initially given in mmHg, and we must convert it to atmospheres for consistency with the gas constant. Use the conversion 1 atm = 760 mmHg:\[7.6 \text{ mmHg} = \frac{7.6}{760} \text{ atm}\]\[\approx 0.01 \text{ atm}\]
03

Rearrange the Equation to Solve for the Number of Moles

Rewrite the Van't Hoff equation to solve for the number of moles \(n\):\[\pi V = nRT \\Rightarrow n = \frac{\pi V}{RT}\]
04

Plug in the Known Values to Find Number of Moles

Substitute the known values into the equation:- \(\pi = 0.01 \text{ atm}\)- \(V = 1.0 \text{ L}\)- \(R = 0.0821\, \text{L atm K}^{-1} \text{mol}^{-1}\)- \(T = 298.15 \text{ K}\)\[n = \frac{(0.01 \text{ atm})(1.0 \text{ L})}{(0.0821\, \text{L atm K}^{-1} \text{mol}^{-1})(298.15 \text{ K})}\]\[\approx 4.08 \times 10^{-4} \text{ moles}\]
05

Calculate Molar Mass with Mass and Moles

The molar mass \(M\) can be calculated using the mass of the polymer and the number of moles:\[M = \frac{\text{mass}}{n}\]where the mass of the polymer is 5.0 g:\[M = \frac{5.0 \text{ g}}{4.08 \times 10^{-4} \text{ moles}}\]\[\approx 12255 \text{ g/mol}\]

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Key Concepts

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

Van't Hoff Equation
The Van’t Hoff equation is essential for understanding osmotic pressure. It allows us to calculate the pressure exerted by solute particles in a solution that has partially permeable membranes. The formula for this is \(\pi = \frac{n}{V}RT\). Here, \(\pi\) represents the osmotic pressure, \(n\) is the number of moles of solute, \(V\) is the volume of the solution (in liters), \(R\) stands for the gas constant \(0.0821\, \text{L atm K}^{-1} \text{mol}^{-1}\), and \(T\) is the temperature in Kelvin.

Using the Van’t Hoff equation lets us translate the osmotic pressure data into valuable molecular insights. It is particularly useful for polymers where other methods might not be as effective.

To ensure results are accurate, it is crucial to use consistent units of measurement. For example, convert osmotic pressure from mmHg to atm to align with the universal gas constant's units. Understanding how each component of the equation correlates helps in solving a variety of solution-based problems.
Molar Mass Calculation
Calculating molar mass becomes straightforward once you know the number of moles from the Van’t Hoff equation. Molar mass \(M\), is found using the formula \(M = \frac{\text{mass}}{n}\). Here, \(\text{mass}\) is the gram weight of the polymer and \(n\) is the number of moles calculated previously.

In this exercise, substituting the given values helps us compute the molar mass of the polymer. With an osmotic pressure of 7.6 mmHg converted to 0.01 atm, and a given mass of 5.0 g, the number of moles comes out to approximately \(4.08 \times 10^{-4}\) moles. Subsequently, dividing the mass by the number of moles yields an approximate molar mass of \(12255 \text{ g/mol}\).

Molar mass calculation is a powerful technique in polymer chemistry, allowing researchers to determine the average molecular weight of complex molecules. It provides essential information about a polymer's physical properties and its potential applications.
Polymer Chemistry
Polymer chemistry involves studying large, complex molecules formed by bonding repeating smaller units known as monomers. The determination of molecular weight or molar mass is critical for understanding how these large molecules, or polymers, will behave under different conditions.

Polymers have a wide variety of applications depending on their molecular structure and molar mass. They are used in everything from plastics to biopolymers crucial in medical applications. Knowing the molar mass helps chemists to tailor and predict the polymer's properties, such as strength, flexibility, and durability.

Within the context of this exercise, determining the molar mass of a polymer using methods like osmotic pressure provides crucial insights into its size and composition.
  • Low molar mass polymers often result in flexible, easily moldable substances.
  • High molar mass polymers tend to be stronger and more durable.

Polymer chemistry is a dynamic field that continues to innovate and enrich our technological landscape.

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Most popular questions from this chapter

Criticize these statements. (a) A saturated solution is always a concentrated one. (b) A \(0.10-\mathrm{mol} / \mathrm{kg}\) sucrose solution and a \(0.10-\mathrm{mol} / \mathrm{kg} \mathrm{KCl}\) solution have the same osmotic pressure.

Calculate the boiling point and the freezing point of these solutions at \(760 \mathrm{mmHg}\). (a) \(20.0 \mathrm{~g}\) citric acid, \(\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{7}\), in \(100.0 \mathrm{~g}\) water (b) \(3.00 \mathrm{~g} \mathrm{CH}_{3} \mathrm{I}\) in \(20.0 \mathrm{~g}\) benzene \(\left(K_{\mathrm{b}}\right.\) benzene \(=\) \(2.53^{\circ} \mathrm{C} \mathrm{kg} / \mathrm{mol} ; K_{\mathrm{f}} \text { benzene } \left.=-5.10^{\circ} \mathrm{C} \mathrm{kg} / \mathrm{mol}\right)\)

The maximum contamination level (MCL) for chlordane is 0.002 ppm. A sample of well water contained 5 ppb chlordane. Is the sample within the MCL for chlordane?

A sample of water contains 0.010 ppm lead ions, \(\mathrm{Pb}^{2+}\). (a) Calculate the mass of lead ions per liter in this solution. (Assume the density of the water solution is \(1.0 \mathrm{~g} / \mathrm{mL}\).) (b) Calculate the mass fraction of lead in ppb.

Calculate the molarity of the solute in a solution containing (a) \(6.18 \mathrm{~g} \mathrm{MgNH}_{4} \mathrm{PO}_{4}\) in \(250 . \mathrm{mL}\) solution. (b) \(16.8 \mathrm{~g} \mathrm{NaCH}_{3} \mathrm{COO}\) in \(300 . \mathrm{mL}\) solution. (c) \(2.50 \mathrm{~g} \mathrm{CaC}_{2} \mathrm{O}_{4}\) in \(750 . \mathrm{mL}\) solution. (d) \(2.20 \mathrm{~g}\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\) in \(400 . \mathrm{mL}\) solution.

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