Question

A 0.15 -g sample of a purified protein is dissolved in water to give 2.0 $\mathrm{mL}$ of solution. The osmotic pressure is found to be 18.6 torr at ${25}^{\circ }\mathrm{C}$ . Calculate the protein's molar mass.

Short answer

The molar mass of the protein is approximately 372 g/mol.

Step-by-step solution

Convert the given data to appropriate units

We need to make sure all the given data are in the correct units. The osmotic pressure is given in torr, which we will need to convert to atm, and the temperature is given in Celsius, which we need to convert to Kelvin. Osmotic pressure in atm: $18.6\frac{torr}{1}\times \frac{1atm}{760torr}=0.0245atm$ Temperature in Kelvin: ${25}^{\circ }C+273.15=298.15K$ Volume in liters: $2.0mL=0.002L$

Rearrange the osmotic pressure equation to find moles

We need to find the number of moles of the protein in the solution. Rearranging the formula for osmotic pressure, we get: $n=\frac{\pi V}{RT}$ Use the converted data from step 1 and solve for the number of moles: $n=\frac{0.0245atm\times 0.002L}{(0.0821Latm/molK)(298.15K)}=0.000403mol$

Calculate the molar mass of the protein

The molar mass of the protein can be calculated by dividing its mass by the number of moles: Molar mass = $\frac{mass}{moles}$ Molar mass = $\frac{0.15g}{0.000403mol}=372g/mol$ The molar mass of the protein is approximately 372 g/mol.

Key concepts

Molar Mass Calculation

Molar Mass Calculation is a fundamental concept when you delve into chemistry. It helps you determine how much a given compound weighs per mole. In our example, we calculated the molar mass of a protein using data from a real-world problem. The process is straightforward once you get the hang of it. Here's a quick guide to understanding this concept better:

• Start by determining the number of moles of the solute. This often involves manipulation of the problem's constraints, like we did with the osmotic pressure formula $\pi V=nRT$.
• After finding the number of moles, use the relationship for molar mass: $\text{Molar Mass}=\frac{\text{mass (grams)}}{\text{moles}}$.
• In this case, the molar mass turned out to be 372 g/mol, meaning each mole of this protein weighs around 372 grams.

By obtaining the molar mass, you gain crucial insight into the scale of microscopic components like proteins.

Protein Chemistry

Protein Chemistry revolves around understanding the diverse structures and functions of proteins, which are complex molecules central to life. In the problem we tackled, calculating the molar mass of a purified protein was part of understanding its properties. Proteins are enormous, made up of long chains of amino acids, and this can impact how you calculate their molar mass. Here's why protein chemistry is fascinating in the context of this example:

• A protein's molar mass helps scientists understand its structure and function. Larger proteins have larger molar masses.
• Proteins can be dissolved and manipulated in solutions to study their properties, as we did with the 0.15 g sample in this exercise.
• Knowledge of a protein's molar mass also assists in applications involving the biological functions they perform, such as catalysis and molecular transport.

Appreciating these aspects is crucial, as proteins are vital for numerous biological processes. By studying their chemistry, we unlock the mysteries of biological systems.

Solution Chemistry

Solution Chemistry is a realm of chemistry focused on mixtures where substances are dissolved. In our exercise, the protein was dissolved in water, creating a solution. This not only allowed us to measure osmotic pressure but also showcased the idea that solutions are ideal mediums for chemical reactions.

• Key concepts include understanding solvents and solutes—water being the solvent and protein as the solute.
• Concentration measurements, like molarity, become crucial. Even though not directly calculated here, they underlie the context of measuring properties like osmotic pressure.
• Osmotic pressure itself can tell us a great deal about the particles in a solution. It's a colligative property, meaning it depends on the number of particles present rather than their identity.

This exercise shows how solution chemistry allows us to quantitatively describe chemical phenomena. These insights enable real-world applications such as drug formulation and biological research, making solution chemistry extremely powerful.