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Suppose you had a balloon made of some highly flexible semipermeable membrane. The balloon is filled completely with a 0.2 \(M\) solution of some solute and is submerged in a \(0.1 M\) solution of the same solute:

Short Answer

Expert verified
The osmotic pressure in the system can be calculated using the formula \(Π = CRT\), where \(C\) is the concentration difference, \(R\) is the gas constant, and \(T\) is the temperature in Kelvin. With a concentration difference of \(ΔC = 0.1 M\) and assuming the temperature is room temperature, which is 25°C or 298.15 K, we can calculate the osmotic pressure as \(Π ≈ 2.452 \:atm\).

Step by step solution

01

1. Identifying the given data

We have a balloon filled with a 0.2 M solution of a solute and submerged in a 0.1 M solution of the same solute.
02

2. Understanding osmotic pressure differences

Osmotic pressure differences occur when there is a difference in solute concentration between two solutions on either side of a semipermeable membrane. This difference creates a pressure across the membrane, causing water to move from the area of lower solute concentration to the area of higher solute concentration.
03

3. Using the osmotic pressure formula

The osmotic pressure (\(Π\)) can be calculated using the formula: \(Π = CRT\), where: \(C\) - the difference in concentration (in moles) between the two solutions, \(R\) - the gas constant (0.0821 L atm/(K mol)), \(T\) - the temperature in Kelvin.
04

4. Determine the concentration difference

Since the balloon contains a 0.2 M solution while submerged in a 0.1 M solution, the concentration difference is: \(ΔC = 0.2 - 0.1 = 0.1 M\)
05

5. Convert the temperature to Kelvin

We need the temperature in Kelvin for the formula. Assuming the temperature is room temperature (25°C), we can convert it to Kelvin: \(T_{K} = T_{C} + 273.15 = 25 + 273.15 = 298.15 K\)
06

6. Calculate the osmotic pressure

Now we can plug the values into the osmotic pressure formula to find the osmotic pressure: \(Π = (0.1 M)(0.0821 \frac{L \cdot atm}{K \cdot mol})(298.15 K)\) \(Π ≈ 2.452 \:atm\) So, the osmotic pressure in this system is approximately 2.452 atm.

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Key Concepts

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

Semipermeable Membrane
A semipermeable membrane is a special type of membrane that allows certain molecules to pass through it while it blocks others. Think of it like a very selective doorway that only lets specific people in. In the context of osmotic pressure, this membrane plays a crucial role.
  • It allows the solvent (often water) to pass, but usually not the solute (the dissolved substance).
  • This selective passage is what enables the process of osmosis, where the solvent naturally moves to balance solute concentrations on either side of the membrane.
  • The movement happens from the side with a low concentration of solute to the side with a higher concentration.
In our balloon example, the flexible membrane lets water move into the balloon, trying to equalize solute concentrations on both sides, but does not allow the solute to pass through. This results in the creation of osmotic pressure that can cause the balloon to expand.
Concentration Gradient
A concentration gradient occurs when there is a difference in solute concentration between two regions. It's like a slope, where the solute wants to "roll down" from a higher concentration to a lower one. This gradient is the driving force for osmosis through a semipermeable membrane.
  • The solution with a higher solute concentration will exert an attraction on water, making it move across the membrane.
  • In the example of the balloon, the 0.2 M solution inside the balloon creates a concentration gradient compared to the 0.1 M solution outside.
  • This gradient is vital because it triggers the movement of water into the balloon, elevating the internal pressure.
Balancing this gradient would mean equalizing the solute's distribution across the membrane, but the existence of the semipermeable membrane means full balancing never quite happens, resulting in constant osmotic pressure.
Gas Constant
The gas constant, often symbolized as \( R \), is a constant used in various fundamental equations in chemistry and physics, including the ideal gas law and osmotic pressure calculation. In osmotic pressure, it links the pressure to the concentration difference and temperature.
  • The value of the gas constant \( R \) is 0.0821 L atm/(K mol).
  • It serves as a conversion factor that helps relate pressure (in atmospheres) with concentration (in moles) and temperature (in Kelvin).
  • By using this constant in the equation \( Π = CRT \), where \( Π \) is osmotic pressure, you can determine how pressure will change with varying concentration and temperature.
In our balloon scenario, \( R \) helped to calculate that the osmotic pressure generated by the concentration gradient at room temperature is approximately 2.452 atm. This calculation illustrates how the gas constant is essential in connecting theoretical osmotic concepts to real-world phenomena.

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Most popular questions from this chapter

A solution contains \(0.115 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}\) and an unknown number of moles of sodium chloride. The vapor pressure of the solution at \(30^{\circ} \mathrm{C}\) is \(25.7\) torr. The vapor pressure of pure water at this temperature is \(31.8\) torr. Calculate the number of moles of sodium chloride in the solution. (Hint: remember that sodium chloride is a strong electrolyte.)

List the following aqueous solutions in order of decreasing freezing point: \(0.040 \mathrm{~m}\) glycerin \(\left(\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}_{3}\right), 0.020 \mathrm{~m}\) \(\mathrm{KBr}, 0.030 \mathrm{~m}\) phenol \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}\right)\).

At \(20^{\circ} \mathrm{C}\) the vapor pressure of benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) is 75 torr, and that of toluene \(\left(\mathrm{C}_{7} \mathrm{H}_{8}\right)\) is 22 torr. Assume that benzene and toluene form an ideal solution. (a) What is the composition in mole fractions of a solution that has a vapor pressure of 35 torr at \(20^{\circ} \mathrm{C} ?\) (b) What is the mole fraction of benzene in the vapor above the solution described in part (a)?

At ordinary body temperature \(\left(37^{\circ} \mathrm{C}\right)\) the solubility of \(\mathrm{N}_{2}\) in water in contact with air at ordinary atmospheric pressure \((1.0 \mathrm{~atm})\) is \(0.015 \mathrm{~g} / \mathrm{L}\). Air is approximately \(78 \mathrm{~mol} \% \mathrm{~N}_{2}\). Calculate the number of moles of \(\mathrm{N}_{2}\) dissolved per liter of blood, which is essentially an aqueous solution. At a depth of \(100 \mathrm{ft}\) in water, the pressure is \(4.0 \mathrm{~atm}\). What is the solubility of \(\mathrm{N}_{2}\) from air in blood at this pressure? If a scuba diver suddenly surfaces from this depth, how many milliliters of \(\mathrm{N}_{2}\) gas, in the form of tiny bubbles, are released into the bloodstream from each liter of blood?

Calculate the molality of each of the following solutions: (a) \(8.66 \mathrm{~g}\) benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) dissolved in \(23.6 \mathrm{~g}\) carbon tetrachloride \(\left(\mathrm{CCl}_{4}\right)\), (b) \(4.80 \mathrm{~g} \mathrm{NaCl}\) dissolved in \(0.350 \mathrm{~L}\) of water.

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