A solution containing \(0.8330 \mathrm{~g}\) of a polymer of unknown structure in \(170.0 \mathrm{~mL}\) of an organic solvent was found to have an osmotic pressure of \(5.20 \mathrm{mmHg}\) at \(25^{\circ} \mathrm{C}\). Determine the molar mass of the polymer.

Short Answer

Expert verified
The molar mass of the polymer is \(175.71 g/mol\).

Step by step solution

01

Convert temperature to Kelvin

The temperature is given in Celsius, and the Van't Hoff formula requires the temperature to be in Kelvin. The conversion is given by \(K = C + 273.15\), so \(T = 25^{\circ} C + 273.15 = 298.15 K\).
02

Convert osmotic pressure to Pascal

The osmotic pressure is given in mmHg, while the gas constant \(R\) value commonly used in the Van't Hoff formula is in Pascals (\(Pa\)). The conversion is \(1 mmHg = 133.322 Pa\), so \(\Pi = 5.20 mmHg * 133.322 Pa/mmHg = 692.87 Pa\).
03

Calculate molar concentration

We can rearrange the Van't Hoff formula to solve for molar concentration \(c\). This gives \(c = \Pi / (RT)\), where \(R = 8.31 J/(mol*K)\) is the gas constant. Substitution gives \(c = 692.87 Pa / (8.31 J/(mol*K) * 298.15 K) = 0.028 mol/L\).
04

Calculate molar mass

The molar mass \(M\) of the polymer can be calculated from the molar concentration and the mass of the polymer in the solution. We have \(M = mass / (volume * concentration)\), where mass is \(0.8330 g\), volume is \(170.0 mL = 0.170 L\), and the concentration is \(0.028 mol/L\). Substitution gives \(M = 0.8330 g / (0.170 L * 0.028 mol/L) = 175.71 g/mol\).

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Key Concepts

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

Osmotic Pressure Calculation
Osmotic pressure is a fundamental concept related to the movement of solvent molecules through a semipermeable membrane, which separates a solution with solute particles from pure solvent. This process occurs until equilibrium is reached. The pressure required to stop this movement and achieve equilibrium is known as the osmotic pressure. It can be experimentally determined and provides valuable information about the properties of the solute, such as its concentration and molar mass.

To calculate osmotic pressure, you can use a simple formula: \( \Pi = iCRT \) where \( \Pi \) is the osmotic pressure, \( i \) is the van't Hoff factor (which indicates the number of particles the solute dissociates into), \( C \) is the molar concentration of the solute, \( R \) is the gas constant, and \( T \) is the temperature in Kelvin. For nonelectrolytes, which do not dissociate in solution, \( i \) is typically 1. The unit of \( \Pi \) often needs converting to be consistent with other units in the formula, such as in the given exercise where we convert from mmHg to Pascals.
Van't Hoff Formula
The Van't Hoff formula, which is used in osmotic pressure calculation, is derived from the ideal gas law. This formula establishes a relationship between osmotic pressure, molar concentration, gas constant, and absolute temperature. In essence, it treats osmotic pressure as if it were an equivalent to the pressure exerted by a gas.

The formula is given by \( \Pi = n/VRT \) or more commonly \( \Pi = CRT \) where \( n \) represents moles of solute, \( V \) the volume of solution, \( C \) the concentration, \( R \) the gas constant (8.31 J/(mol*K)), and \( T \) the absolute temperature in Kelvin. It is important to notice that all elements of the equation must be in proper units to ensure a correct calculation. For instance, converting Celsius to Kelvin for temperature and mmHg to Pascals for pressure, as done in the solution, is essential for accuracy.
Molar Mass from Osmotic Pressure
The molar mass of a solute can be calculated from the osmotic pressure of its solution. This is a practical application of the Van't Hoff equation, which offers a method to determine the molar mass without knowing the chemical structure of the solute.

Once you have the osmotic pressure, you can rearrange the equations to solve for the molar concentration \( C \) of the solute, which you then use to find the molar mass \( M \) using the relation \( M = \frac{mass}{volume \times concentration} \). In this approach, the mass of the solute and the volume of the solution are taken from the experimental data. This step-by-step process identifies the molar mass that characterizes the unknown solute. The resulting molar mass has extensive applications, including the study of molecular structures, polymer science, and in fields where molecular size is a critical parameter.

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Most popular questions from this chapter

For ideal solutions, the volumes are additive. This means that if \(5 \mathrm{~mL}\) of \(\mathrm{A}\) and \(5 \mathrm{~mL}\) of \(\mathrm{B}\) form an ideal solution, the volume of the solution is \(10 \mathrm{~mL}\). Provide a molecular interpretation for this observation. When \(500 \mathrm{~mL}\) of ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) are mixed with \(500 \mathrm{~mL}\) of water, the final volume is less than \(1000 \mathrm{~mL}\). Why?

Calculate the percent by mass of the solute in each of the following aqueous solutions: (a) \(5.50 \mathrm{~g}\) of \(\mathrm{NaBr}\) in \(78.2 \mathrm{~g}\) of solution, (b) \(31.0 \mathrm{~g}\) of \(\mathrm{KCl}\) in \(152 \mathrm{~g}\) of water, (c) \(4.5 \mathrm{~g}\) of toluene in \(29 \mathrm{~g}\) of benzene.

Why are ice cubes (for example, those you see in the trays in the freezer of a refrigerator) cloudy inside?

The Henry's law constant of oxygen in water at \(25^{\circ} \mathrm{C}\) is \(1.3 \times 10^{-3} \mathrm{~mol} / \mathrm{L} \cdot\) atm. Calculate the molarity of oxygen in water under 1 atmosphere of air. Comment on the prospect for our survival without hemoglobin molecules. (Recall from previous problems that the total volume of blood in an adult human is about \(5 \mathrm{~L}\).)

A mixture of ethanol and 1-propanol behaves ideally at \(36^{\circ} \mathrm{C}\) and is in equilibrium with its vapor. If the mole fraction of ethanol in the solution is \(0.62,\) calculate its mole fraction in the vapor phase at this temperature. (The vapor pressures of pure ethanol and 1 -propanol at \(36^{\circ} \mathrm{C}\) are \(108 \mathrm{mmHg}\) and \(40.0 \mathrm{mmHg}\), respectively.

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