A \(0.10-\mathrm{g}\) sample of a polymer, dissolved in \(100 \mathrm{~mL}\). of toluene, had an osmotic pressure of \(5.4\) Torr at \(20^{\circ} \mathrm{C}\). What is the molar mass of the polymer?

Short Answer

Expert verified
The molar mass of the polymer is calculated using the number of moles obtained from the osmotic pressure equation and the given mass of the polymer.

Step by step solution

01

Understanding osmotic pressure

Osmotic pressure (\( \text{Π} \)) can be calculated using the formula \( \text{Π} = \text{nRT}/\text{V} \), where \( \text{n} \) is the number of moles of the solute, \( \text{R} \) is the gas constant (0.0821 L·atm/(K·mol)), \( \text{T} \) is the temperature in Kelvin, and \( \text{V} \) is the volume of the solution in liters.
02

Convert temperature to Kelvin

The temperature must be in Kelvin for the osmotic pressure equation. Convert \( 20^\circ \text{C} \) to Kelvin by adding \( 273 \). Thus the temperature in Kelvin is \(\text{T} = 20 + 273 = 293 \text{ K}\).
03

Convert osmotic pressure to atmospheres

Convert the osmotic pressure from Torr to atmospheres, as the gas constant \( \text{R} \) is given in these units. Use the conversion \(\text{1 atm} = 760 \text{ Torr}\). Therefore, \( \text{Π} = 5.4 \text{ Torr} \) is equal to \( \frac{5.4}{760} = 0.007105 \text{ atm}\).
04

Convert volume to liters

The volume of the solution should be in liters. Convert \(100 \text{ mL} \) to liters by dividing by \(1000 \text{ mL/L}\). Thus the volume in liters is \( \text{V} = 0.100 \text{ L}\).
05

Solve for the number of moles (\text{n})

Rearrange the osmotic pressure equation to solve for \( \text{n} \): \(\text{n} = \text{ΠV}/\text{RT}\). Substitute the known values into the equation: \(\text{n} = (0.007105 \text{ atm} \times 0.100 \text{ L}) / (0.0821 \text{ L·atm/(K·mol)} \times 293 \text{ K})\). Solve for \( \text{n} \).
06

Calculate the molar mass of the polymer

Molar mass (\text{M}) is the mass of the polymer divided by the number of moles calculated. Use the formula \( \text{M} = \text{mass}/\text{n} \). The mass of the polymer sample is \(0.10 \text{ g}\). Plug in the values to get \( \text{M} \): \(\text{M} = 0.10 \text{ g} / \text{n}\). Once \( \text{n} \) is calculated from Step 5, compute \( \text{M} \).

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Key Concepts

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

Molar Mass Determination
Molar mass is a fundamental concept in chemistry that represents the mass of one mole of a substance. It is crucial to determining the molecular weight of a compound, which can then be used to calculate the number of moles in a given sample. In the context of the exercise, we are interested in finding the molar mass of a polymer.

To find the molar mass, we first need to understand the relationship between mass, molar mass, and the number of moles. The general formula is:
\[ M = \frac{mass}{n} \]
Where \( M \) is the molar mass, mass is the given mass of the substance, and \( n \) is the number of moles.

In practice, to determine the molar mass of the polymer from the exercise, we calculate the number of moles using the known osmotic pressure, volume, and temperature with the formula for osmotic pressure, and then we use the above formula to find the molar mass. This approach is typical in polymer chemistry where direct measurement might be challenging due to the large size and complexity of the molecules.
Colligative Properties
Colligative properties are properties of solutions that depend on the ratio of the number of solute particles to the solvent molecules and not on the nature of the chemical species present. These properties include boiling point elevation, freezing point depression, vapor pressure lowering, and osmotic pressure.

Osmotic pressure is a colligative property that is particularly important for molar mass determination of large molecules such as polymers. It is the pressure required to stop the flow of solvent molecules through a semipermeable membrane from a region of low solute concentration to one of high concentration.

The formula to calculate osmotic pressure is given by: \[\Pi = \frac{nRT}{V}\]
where \( \Pi \) symbolizes the osmotic pressure, \( n \) represents the number of moles of solute, \( R \) is the ideal gas constant, \( T \) is the temperature in Kelvin, and \( V \) is the volume of the solution.

Understanding the concept of osmotic pressure is critical, as it applies to various disciplines, including biology (e.g., plant water uptake and kidney function) and materials science (e.g., membrane technologies).
Polymer Chemistry
Polymer chemistry is a branch of chemistry that focuses on the chemical synthesis and chemical properties of polymers or macromolecules. Polymers consist of repeating structural units called monomers, and their large size makes their physical properties significantly different from those of small molecules.

Aspects of Polymer Behavior

Polymers exhibit unique behaviors, such as viscoelasticity and the ability to form more complex structures like block copolymers. Their properties can be tailored for specific uses, ranging from everyday products like plastic containers to high-performance materials in aerospace.

In the context of the textbook problem, the polymer solution's osmotic pressure can provide insights into the polymer's molar mass, which is crucial for understanding its physical and chemical properties. Due to the large size of polymer molecules, typical methods of molar mass determination, such as freezing point depression, are often not viable. That's why osmotic pressure measurements are particularly useful in polymer chemistry for molar mass determination. Moreover, the study of polymers as a subject intertwines physics, engineering, and materials science, making it an interdisciplinary and dynamic field.

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Most popular questions from this chapter

The enthalpy of solution for ammonium nitrate in water is positive. (a) Does \(\mathrm{NH}_{4} \mathrm{NO}_{3}\) dissolve endothermically or exothermically? (b) Write the chemical equation for the dissolving process. (c) Which is larger for \(\mathrm{NH}_{4} \mathrm{NO}_{3}\), the lattice enthalpy or the enthalpy of hydration?

Explain in thermodynamic terms why a solute raises the boiling point and lowers the freceing point of a solution.

The carbon dioxide gas dissolved in a sample of water in a partially filled, sealed containcr has reached equilibrium with its partial pressure in the air above the solution. Explain what happens to the solubility of the \(\mathrm{CO}_{2}\) if (a) the partial pressure of the \(\mathrm{CO}_{2}\) gas is doubled by the addition of more \(\mathrm{CO}_{2} ;\) (b) the total pressure of the gas above the liquid is doubled by the addition of nitrogen.

What is meant by dynamic equilibrium? How does it differ from static equilibrium?

A \(0.020 \mathrm{M} \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(\mathrm{aq})\) solution is separated from a \(0.050 \mathrm{M} \mathrm{CO}\left(\mathrm{NH}_{2}\right)_{2}(\mathrm{aq})\) solution by a semipermeable membrante at \(25^{\circ} \mathrm{C}\). (a) Which solution has the higher osmotic pressure? (b) Which solution becomes more dilute with the passage of \(\mathrm{H}_{2} \mathrm{O}\) molecules through the membrane? (c) To which solution should an external pressure be applied in order to maintain an equilibrium flow of \(\mathrm{H}_{2} \mathrm{O}\) molecules across the membrane? (d) What external pressure (in atm) should be applied in (c)?

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