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The protein human plasma, albumin, has a molecular weight of \(69,000 .\) Calculate the osmotic pressure of a solution of this protein containing \(2 \mathrm{~g}\) per \(100 \mathrm{~cm}^{3}\) at \(25^{\circ} \mathrm{C}\) in (a) Torr and (b) millimeters of water. The experiment is carried out using a salt solution for solvent and a membrane permeable to salt.

Short Answer

Expert verified
The osmotic pressure of the human plasma albumin solution is approximately \(55.56\) Torr and \(729.14\) mm of water.

Step by step solution

01

Convert temperature to Kelvin

To work with the formula, we need to convert the temperature from Celsius to Kelvin. To do this, we add 273.15 to the Celsius temperature. T (K) = 25°C + 273.15 = 298.15 K
02

Determine the volume of the solution in liters

We are given a volume of 100 cm^3. Let's convert this to liters for the calculations. 1 cm^3 = 0.001 L, so 100 cm^3 = 0.100 L
03

Find the moles of solute in the solution

To determine the number of moles of solute (protein), divide the mass of the solute by its molecular weight. n = (2 g) / (69,000 g/mol) = 2.8985 x 10^{-5} mol
04

Calculate the osmotic pressure in atmospheres

Now that we have all the values, we can use the formula for osmotic pressure (π = i * n/V * R * T). π = (1) * (2.8985 x 10^{-5} mol)/(0.100 L) * (0.0821 L atm K^{-1} mol^{-1}) * (298.15 K) π = 0.0731 atm
05

Convert osmotic pressure to Torr and mm of water

Finally, we need to convert the osmotic pressure from atmospheres to Torr and millimeters of water. (a) To convert to Torr, use the conversion factor 1 atm = 760 Torr. Osmotic pressure in Torr = (0.0731 atm)*(760 Torr/atm) = 55.56 Torr (b) To convert to millimeters of water, use the conversion factor 1 atm = 101325 Pa (Pascal) and 1 Pa = 9.81 x 10^{-3} mm of water. Osmotic pressure in water(mm) = (0.0731 atm)*(101325 Pa/atm)*(9.81 x 10^{-3} mm of water/Pa) = 729.14 mm of water Thus, the osmotic pressure of the solution is approximately 55.56 Torr and 729.14 mm of water.

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Key Concepts

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

Molecular Weight
Understanding molecular weight is crucial when studying chemical substances, as it represents the mass of a single molecule, usually expressed in unified atomic mass units or Daltons (Da). When we talk about proteins like human plasma albumin, the molecular weight becomes exceedingly high, reaching tens of thousands or more Da.

For calculations in chemistry or biochemistry, the molecular weight is commonly converted to grams per mole (g/mol), which describes the mass of one mole of the substance. This conversion allows for the use of the ideal gas law and related equations. Knowing the molecular weight of albumin, as given in the exercise, enables us to calculate the number of moles in the solution, which is a critical step towards determining the osmotic pressure.
Solution Concentration
The concentration of a solution refers to how much solute is dissolved within a solvent. It can be expressed in various ways, such as mass per unit volume, molarity, or molality. In this exercise, the solution concentration is given as grams per cubic centimeter.

To link this to other concentration measures for our calculations, we need to relate the mass of the solute to the volume of the solvent. The concentration helps us figure out how many moles of solute are present in a given volume of solvent, which is crucial for calculating osmotic pressure.
Temperature Conversion
Temperature plays a vital role in various scientific calculations, with the Kelvin scale being the standard unit of temperature in physical sciences. To convert Celsius to Kelvin, you add 273.15 to the Celsius temperature, allowing for the use of the formula for osmotic pressure. This standardization is essential because the Kelvin scale provides an absolute scale where zero (0 K) corresponds to absolute zero, the theoretical condition where molecules cease to move.
van't Hoff Factor
The van't Hoff factor (i) accounts for the number of particles a solute dissociates into when it dissolves. For non-electrolytes, such as the protein in our example, i is typically considered to be 1 because they do not dissociate into ions in solution. Understanding the van't Hoff factor is crucial, as it can significantly influence the colligative properties of solutions such as osmotic pressure, boiling point elevation, and freezing point depression.
Molarity
Molarity is one of the ways to express the concentration of a solution, defined as the number of moles of solute per liter of solution. It is a unit of concentration commonly used in chemistry for stochiometric calculations, titrations, and preparing solutions. In the context of osmotic pressure calculation, molarity directly relates the amount of solute to the volume of the solution, forming an essential part of the equation utilized.
Ideal Gas Law
The ideal gas law is a fundamental equation in chemistry that relates pressure, volume, number of moles, and temperature of a gas in a simplistic way. While it’s derived for gases, osmotic pressure calculations employ a similar formula as solutions can often be treated analogously to gases in a dilute system. For our problem, we use the equation for osmotic pressure (π = i * n/V * R * T), which parallels the ideal gas law and allows us to calculate the pressure exerted by the dissolved solute in the solution.
Pressure Units Conversion
When dealing with pressure, there are numerous units we might encounter, including atmospheres (atm), Torr, pascals (Pa), and millimeters of water (mm H2O). Converting between these units is often necessary in scientific calculations to report results in a desired unit of measure. Each unit has its conversion factor: 1 atm equals 760 Torr or 101325 Pa, and these factors are used to express the osmotic pressure in the units required by different contexts or scientific standards.

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Most popular questions from this chapter

The sedimentation and diffusion coefficients for hemoglobin corrected to \(20^{\circ}\) in water are \(4.41 \times 10^{-13} \mathrm{sec}\) and \(\left(6.3 \times 10^{-11} \mathrm{~m}^{2} / \mathrm{s}\right)\), respectively. If \(\underline{\mathrm{V}}=\left(.749 \mathrm{~cm}^{3} / \mathrm{g}\right)\) and \(\mathrm{P}_{(\mathrm{H})} 20=\left(0.998 \mathrm{~g} / \mathrm{cm}^{3}\right)\) at this temperature, calculate the molecular weight of the protein. If there is \(1 \mathrm{~g}\) -atom of iron per \(17,000 \mathrm{~g}\) of protein, how many atoms of iron are there per hemoglobin molecule?

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