Problem 78 - Chemistry: The Central Science

Question

A dilute aqueous solution of an organic compound soluble in water is formed by dissolving \(2.35 \mathrm{~g}\) of the compound in water to form \(0.250 \mathrm{~L}\) solution. The resulting solution has an osmotic pressure of \(0.605 \mathrm{~atm}\) at \(25^{\circ} \mathrm{C}\). Assuming that the organic compound is a nonelectrolyte, what is its molar mass?

Step-by-Step Solution

Answer

The molar mass of the organic compound is approximately 376 g/mol.

Step 1: Convert temperature to Kelvin

First, we need to convert the temperature from Celsius to Kelvin. The conversion formula is: \[ T_{(K)} = T_{(°C)} + 273.15 \] Given the temperature in Celsius, we can find the temperature in Kelvin: \[ T_{(K)} = 25°C + 273.15 = 298.15 \mathrm{K} \]

Step 2: Calculate molarity from osmotic pressure equation

The osmotic pressure equation is: \[ \pi = CRT \] We need to solve for the molarity, \(C\): \[ C = \frac{\pi}{RT} \] Given the osmotic pressure \(\pi = 0.605 \mathrm{~atm}\), gas constant \(R = 0.0821 \mathrm{L~atm/mol~K}\), and temperature in Kelvin, we can calculate the molarity: \[ C = \frac{0.605 \mathrm{~atm}}{(0.0821 \mathrm{L~atm/mol~K})(298.15 \mathrm{K})} \approx 0.0250 \mathrm{M} \]

Step 3: Calculate the moles of the organic compound

Using the molarity and volume of the solution, we can find the moles of the organic compound: \[ moles = Molarity \times Volume \] Given the molarity and the volume of the solution, we can calculate the moles of the organic compound: \[ moles = (0.0250 \mathrm{M})(0.250 \mathrm{L}) = 0.00625 \mathrm{mol} \]

Step 4: Calculate the molar mass of the organic compound

Using the mass and moles, we can find the molar mass of the organic compound: \[ Molar \hspace{1mm} Mass = \frac{Mass}{Moles} \] Given the mass of the organic compound and the moles, we can calculate the molar mass: \[ Molar \hspace{1mm} Mass = \frac{2.35 \mathrm{g}}{0.00625 \mathrm{mol}} \approx 376 \mathrm{g/mol} \] Therefore, the molar mass of the organic compound is approximately 376 g/mol.

Key Concepts

Molar Mass Determination

Understanding the molar mass of compounds is fundamental in chemistry, especially when working with solutions. To determine the molar mass, we often use the mass of a sample and the amount of substance in moles contained within that sample.

The key equation is:
\[ Molar \ Mass = \frac{Mass}{Moles} \]
In our example, after finding the moles of the compound with the help of osmotic pressure, we divide the mass of the compound in grams by the number of moles to get the molar mass. This technique is not only crucial for academic purposes but also for several industrial applications such as pharmaceuticals where precise dosages of medication depend on accurate molar mass measurements.

Colligative Properties

Colligative properties are characteristics of solutions that depend on the concentration of dissolved particles but not on their identity. Osmotic pressure is a significant colligative property, especially in biological and chemical processes.

Osmotic Pressure as a Colligative Property

Osmotic pressure is the pressure required to stop the flow of solvent molecules through a semipermeable membrane from a dilute solution to a more concentrated solution. It is directly proportional to the molarity of the solution, which implies the more solute particles, the higher the osmotic pressure. Osmotic pressure can be harnessed to calculate the molar mass of unknown substances when they're dissolved in solution.

Solution Concentration

Solution concentration is a measure of the amount of solute present in a given quantity of solvent or solution. In the realm of osmotic pressure, concentration is usually expressed as molarity (M), calculated as moles of solute per liter of solution.

Calculating Molarity from Osmotic Pressure

To find the molarity from osmotic pressure, the formula used is an adaptation of the ideal gas law equation:
\[ C = \frac{\pi}{RT} \]
This equation lets us derive the concentration once we have the osmotic pressure (π), the gas constant (R), and the temperature in Kelvin (T). It's useful in a variety of scientific disciplines, from pharmacology to environmental science, where understanding solution concentration is key to both product formulation and research.

Gas Laws

The gas laws help us understand the behavior of gases under different conditions, and they are also applicable to solutions in certain contexts, like osmotic pressure. Boyle’s, Charles’s, and Avogadro’s laws all merge into the ideal gas equation, which is similar in form to the equation for osmotic pressure.

Applying Gas Laws to Osmotic Pressure

The osmotic pressure equation is analogous to the ideal gas law (PV=nRT), underscoring the relationship between pressure, volume, and temperature. In our context:\[ \pi V = nRT \]
Here, π represents osmotic pressure, V is the volume, n is the number of moles of solute, R is the gas constant, and T is the temperature in Kelvin. This allows us to use concepts from gas laws to solve problems involving concentrations and pressures in solutions.