Problem 25
Question
A protein has a molar mass of \(69000 \mathrm{gmol}^{-1}\). Assuming ideal behaviour, calculate the osmotic pressure of a solution containing \(20 \mathrm{g} \mathrm{dm}^{-3}\) of the protein at \(298 \mathrm{K}\). (Section 17.4 )
Step-by-Step Solution
Verified Answer
The osmotic pressure is approximately 0.00708 atm or 717.5 Pa.
1Step 1: Understand the Osmotic Pressure Formula
The formula for osmotic pressure \( \pi \) of a solution is given by the van't Hoff equation: \[ \pi = \frac{n}{V}RT \]where \( n \) is the number of moles of solute, \( V \) is the volume of the solution in liters, \( R \) is the ideal gas constant (0.0821 L atm mol⁻¹ K⁻¹), and \( T \) is the temperature in Kelvin. Since the concentration \( C \) is given in \( \mathrm{g} \mathrm{dm}^{-3} \), we need to relate this to \( n \) and \( V \).
2Step 2: Convert Mass Concentration to Moles
The mass concentration is \( 20 \mathrm{g} \mathrm{dm}^{-3} \). Convert this to molarity (\( n/V \)) using the molar mass of the protein: \[ \text{Molarity} = \frac{20 \mathrm{g} \mathrm{dm}^{-3}}{69000 \mathrm{gmol}^{-1}} \approx 2.8985 \times 10^{-4} \mathrm{mol} \mathrm{dm}^{-3} \] This is the concentration \( C \) that we will use in our calculations.
3Step 3: Apply the Van't Hoff Equation
Using the Van't Hoff equation \( \pi = CRT \), substitute the known values: \[ C = 2.8985 \times 10^{-4} \mathrm{mol} \mathrm{dm}^{-3}, \ R = 0.0821 \mathrm{L} \mathrm{atm} \mathrm{mol}^{-1} \mathrm{K}^{-1}, \ T = 298 \mathrm{K} \]Thus, \[ \pi = (2.8985 \times 10^{-4}) \times (0.0821) \times (298) \approx 0.00708 \mathrm{atm} \]
4Step 4: Convert Units if Necessary
The osmotic pressure is calculated in atm. If you need it in another unit, like Pascal (1 atm = 101325 Pascal), you can convert it:\[ \pi = 0.00708 \mathrm{atm} \times 101325 \mathrm{Pa/atm} \approx 717.5 \mathrm{Pa} \]
Key Concepts
van't Hoff equationideal gas constantmolarity calculationmolar mass of protein
van't Hoff equation
Osmotic pressure is an important concept in chemistry and biology and can be calculated using the van't Hoff equation. This equation relates the osmotic pressure of a solution to its solute concentration, temperature, and a constant. The formula is expressed as:
- \( \pi = CRT \)
- \( \pi \) is the osmotic pressure.
- \( C \) is the molarity of the solution. This is the concentration expressed in moles of solute per liter of solution.
- \( R \) is the ideal gas constant (explained in detail below).
- \( T \) is the temperature, measured in Kelvin.
ideal gas constant
The ideal gas constant, represented as \( R \), is a crucial factor in various chemical equations, including the van't Hoff equation. It acts as a proportionality factor that relates the other variables in these equations. The value of \( R \) can vary depending on the units used for pressure, volume, and temperature, but in the context of osmotic pressure calculations:
- \( R = 0.0821 \text{ L atm mol}^{-1} \text{ K}^{-1} \)
molarity calculation
Calculating molarity involves determining how many moles of a solute are present in a specific volume of solution. Here's how you can calculate molarity with given mass:
- Start with the mass of the solute you have, which in this exercise was 20 g of protein per liter of solution.
- Use the molar mass of the protein (given as 69000 g/mol) to convert this mass into moles. The formula is:
\[ \text{Molarity} = \frac{\text{mass of solute (g)}}{\text{molar mass (g/mol)}} \]
- Convert 20 g of protein to moles using \( \text{Molarity} \approx \frac{20}{69000} \approx 2.8985 \times 10^{-4} \text{ mol/dm}^3 \)
molar mass of protein
The molar mass of a compound, especially large macromolecules like proteins, is an essential value that indicates how much one mole of that compound weighs. For proteins, this can be used to convert their concentration from grams per liter to moles per liter, which is crucial for calculations involving molarity.
- In this example, the molar mass of the protein was given as 69000 g/mol.
- This implies that one mole of this protein would weigh 69000 grams.
Other exercises in this chapter
Problem 22
The vapour pressure of pure \(\mathrm{CHCl}_{3}\) at \(318 \mathrm{K}\) is \(58 \mathrm{kPa}\). What would be the partial vapour pressure of \(\mathrm{CHCl}_{3}
View solution Problem 23
Ethanol and methanol form nearly ideal solutions. At \(20^{\circ} \mathrm{C}\), the vapour pressure of pure ethanol is 5930 Pa and of methanol is \(11830 \mathr
View solution Problem 27
A solution of iodoethane, \(1,\) and propanone, \(P\), with a mole fraction of \(I, x_{1}=0.55,\) had a partial vapour pressure of I of \(28.44 \mathrm{kPa}\) a
View solution Problem 28
When a patient suffers severe dehydration, why will medical treatment involve giving 'saline' solution rather than pure water? (Section \(17.4)\)
View solution