A solution prepared by adding \(0.50 \mathrm{~g}\) of a polymer to enough toluene (an organic solvent) to make \(200 \mathrm{~mL}\). of solution had an osmotic pressure of \(0.582\) 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 approximately 1710 g/mol.

Step by step solution

01

Recall the osmotic pressure equation

Use the osmotic pressure equation which relates osmotic pressure (\text{symbol: \text{\(\Pi\)}}), molar concentration (C, in moles/L), the gas constant (R), and temperature (T, in Kelvin). The equation is \text{\text{\(\Pi = CRT\)}}.
02

Convert the temperature to Kelvin

Convert the temperature from Celsius to Kelvin by adding 273.15 to the Celsius temperature. The Kelvin temperature is therefore \text{\text{\(T = 20 + 273.15 = 293.15 \text{K}\)}}.
03

Rearrange the equation to solve for molar concentration (C)

Rearrange the osmotic pressure equation to solve for C. Given that \text{\text{\(\Pi = 0.582 \text{Torr}\)}}, convert the osmotic pressure to atmospheres by dividing by 760. The rearranged equation is \text{\text{\(C = \frac{\Pi}{RT}\)}}.
04

Calculate the molar concentration (C)

Plug the osmotic pressure, gas constant (R = 0.0821 L·atm/K·mol), and temperature in Kelvin into the rearranged equation to calculate C. Note that the pressure unit for R is in atmospheres, so first convert \text{\text{\(0.582 \text{Torr}\)}} to atmospheres by dividing by \text{760}.
05

Compute the molar mass (M) of the polymer

Use the concentration and the mass of the polymer to find the molar mass. Since we've found C from Step 4 and we know the mass (\text{\text{\(0.50 \text{g}\)}}), the molar mass M is given by \text{\text{\(M = \frac{\text{mass (g)}}{\text{volume (L)} \times C}\)}}.
06

Convert the solution volume to liters

Convert the volume of the solution from milliliters to liters by dividing by 1000. Therefore, \text{\text{\(200 \text{mL} = 0.200 \text{L}\)}}.
07

Calculate the molar mass (M) of the polymer

Insert the values of the mass of the polymer, the volume of the solution in liters, and the molar concentration into the equation \text{\text{\(M = \frac{\text{mass (g)}}{\text{volume (L)} \times C}\)}} to find the molar mass of the polymer.

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Key Concepts

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

Molar Mass
Understanding molar mass is essential for students studying chemistry, as it represents the mass of one mole of a substance and is expressed in grams per mole (g/mol). In the context of the osmotic pressure calculation from our exercise, molar mass is the key to determining the quantity of the polymer in molecular terms.

To calculate molar mass, you take the mass of the substance and divide it by the number of moles present. In the given exercise, the molar mass (M) is determined by using the formula:
\[\begin{equation} M = \frac{\text{mass (g)}}{\text{volume (L)} \times C} \end{equation}\]
Since you know the mass of the polymer (\text{0.50 g}) and the volume of the solution (\text{0.200 L}), you can find the molar mass once you have calculated the molar concentration (C) from the osmotic pressure. The molar mass is a fundamental concept not only for this problem but for various applications including stoichiometry, preparation of solutions, and conversion of grams to moles in chemical equations.

It's also important to understand that molar mass can be used to identify the molecular formula of an unknown compound by comparing the calculated molar mass to known values, which is particularly useful in organic chemistry.
Van't Hoff Factor
In solutions chemistry, the van't Hoff factor (i) is a dimensionless quantity used to describe the effect of solute dissociation or association on colligative properties, which are properties that depend on the ratio of the number of solute particles to the solvent molecules in a solution. Colligative properties include boiling point elevation, freezing point depression, vapor pressure lowering, and osmotic pressure.

The van't Hoff factor is defined as the ratio of moles of particles in solution to moles of solute dissolved. For non-electrolytes, substances that do not dissociate into ions in solution, the van't Hoff factor is 1. For electrolytes, it can be greater than 1, if the substance dissociates into multiple ions, or less than 1, if it associates into fewer particles.

In the given problem, the polymer is assumed not to dissociate or associate in the solvent toluene, so the van't Hoff factor is implicitly 1. However, for accurate calculations involving ionic compounds, the van't Hoff factor must be considered, as it will affect the molar concentration and therefore the calculated osmotic pressure. When studying solutions and solubility, taking into account the van't Hoff factor is crucial for students to predict and understand the behavior of solutions under various conditions.
Solutions and Solubility
Solutions and solubility are fundamental concepts in chemistry, involving the process by which one substance (the solute) dissolves in another (the solvent) to form a homogenous mixture known as a solution. Solubility is the measure of how much solute can be dissolved in a solvent at a given temperature and pressure.

The solubility of a substance depends on the nature of the solute and solvent, temperature, pressure, and in some cases, the pH of the solution. In our polymer and toluene example, toluene acts as the solvent and the polymer as the solute, and the solubility is expressed in terms of osmotic pressure which relates to how well the polymer dissolves in the solvent.

Factors Affecting Solubility

Several factors influence solubility, and understanding these can help students foresee and control the solubility of substances:
  • Temperature: In most cases, solubility increases with temperature.
  • Pressure: For gases, solubility is directly proportional to pressure.
  • Chemical Nature: 'Like dissolves like', polar solutes dissolve better in polar solvents, and non-polar solutes in non-polar solvents.
  • Intermolecular Forces: Stronger forces between solvent molecules and solute lead to higher solubility.
These factors are not just crucial for predicting the behavior of solutions but also for practical applications such as drug formulation, where solubility can influence the efficacy of a medicinal compound.

Overall, students must grasp these core concepts to understand more complex phenomena such as osmosis and to apply this understanding to calculate osmotic pressure in a variety of contexts.

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