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Chapter 12: Problem 131

A very long pipe is capped at one end with a semipermeable membrane. How deep (in meters) must the pipe be immersed into the sea for freshwater to begin to pass through the membrane? Assume the water to be at \(20^{\circ} \mathrm{C}\) and treat it as a \(0.70 \mathrm{M} \mathrm{NaCl}\) solution. The density of seawater is \(1.03 \mathrm{~g} / \mathrm{cm}^{3}\) and the acceleration due to gravity is \(9.81 \mathrm{~m} / \mathrm{s}^{2}\)

Short Answer

Expert verified
The answer will be the result obtained in Step 4. It should be a numerical value representing the depth the pipe must be immersed for osmosis of fresh water to begin.

Step by step solution

01

Calculate the pressure increase due to osmosis

Using the formula for osmotic pressure, \(\Pi = iMRT\), where \(i\) is the van't Hoff factor (assumed to be 2 for NaCl, which fully dissociates in water), \(M\) is the molarity (0.70 M), \(R\) is the ideal gas law constant (which is 0.0821 L.atm/mol.K in suitable units), and \(T\) is the temperature in Kelvin (293.15 K at 20掳C), we find the osmotic pressure. This physical quantity represents the pressure increase due to osmotic forces.
02

Convert to SI units

Osmotic pressure is effectively a measure of total force exerted by a solution due to solute particles in a given area. The units from Step 1 are in atmospheres and need to be converted to Pascal (Pa), the SI unit for pressure. To do so, use the conversion factor of 1 atm equal to 101325 Pa.
03

Equate to fluid pressure

This osmotic pressure needs to be equated to fluid pressure, since osmosis stops when the osmotic pressure equals the fluid pressure. The formula for fluid pressure is \(\rho g h\), where \(\rho\) is the water density (1.03 g/cm^3, but need to convert into kg/m^3 using conversion factor 1000), \(g\) is the acceleration due to gravity (9.81 m/s^2) and \(h\) is the height of water column in m which we need to find.
04

Solve for h

Solving the equation from step 3 for \(h\), we find the height required for the osmotic pressure to equal the fluid pressure. This will be the height (in meters) the pipe must be immersed in the sea for freshwater to pass through the semipermeable membrane by osmosis.

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Key Concepts

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

Understanding Semipermeable Membranes
A semipermeable membrane is a special type of barrier. It allows certain molecules or ions to pass through it, while blocking others. In the case of the exercise, the membrane at the end of the pipe only allows water molecules to pass, but not the dissolved salts from the sea water.
This type of membrane is crucial for processes like osmosis, where the unequal distribution of solute drives the movement of water. By selectively permitting the passage of solvent, semipermeable membranes help establish different concentrations on each side.
Exploring NaCl Solution
When we refer to an NaCl solution, we're talking about saltwater. In this solution, sodium chloride (NaCl) is fully dissolved in water. This dissociation results in sodium (Na鈦�) and chloride ions (Cl鈦�), which are responsible for the solution's properties.
NaCl solutions are common in environments like sea water. The exercise uses a 0.70 M NaCl solution, indicating it has 0.70 moles of NaCl per liter, influencing the osmotic pressure significantly. This concentration affects how water will move through the semipermeable membrane during osmosis.
The Role of Fluid Pressure
Fluid pressure is the force exerted by a fluid per unit area. It's a key factor in determining how long the pipe needs to be to balance osmotic pressure.
In our scenario, the fluid pressure is calculated using the formula \(\rho g h\), where \(\rho\) is the density of sea water (converted into kg/m鲁), \(g\) is gravity's acceleration, and \(h\) represents the depth of immersion. This pressure opposes osmotic pressure until equilibrium is achieved.
Demystifying Osmosis
Osmosis is a natural process where water moves through a semipermeable membrane from a region of lower solute concentration to a region of higher solute concentration. It's driven by the difference in solute concentrations on either side of the membrane.
In the exercise, as freshwater moves through the membrane, it's attempting to equalize concentrations between the two sides. The osmotic pressure, calculated using the formula \(\Pi = iMRT\), dictates how much force is needed for this water movement. Understanding osmosis helps us determine how much pressure and depth are required for water to pass through the membrane.

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Most popular questions from this chapter

An aqueous solution of a \(0.10 M\) monoprotic acid HA has an osmotic pressure of 3.22 atm at \(25^{\circ} \mathrm{C}\). What is the percent ionization of the acid at this concentration?

A mixture of ethanol and 1-propanol behaves ideally at \(36^{\circ} \mathrm{C}\) and is in equilibrium with its vapor. If the mole fraction of ethanol in the solution is \(0.62,\) calculate its mole fraction in the vapor phase at this temperature. (The vapor pressures of pure ethanol and 1 -propanol at \(36^{\circ} \mathrm{C}\) are \(108 \mathrm{mmHg}\) and \(40.0 \mathrm{mmHg}\), respectively.

Calculate the molalities of the following aqueous solutions: (a) \(1.22 M\) sugar \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) solution (density of solution \(=1.12 \mathrm{~g} / \mathrm{mL}),\) (b) \(0.87 \mathrm{M}\) \(\mathrm{NaOH}\) solution (density of solution \(=1.04 \mathrm{~g} / \mathrm{mL})\) (c) \(5.24 \mathrm{M} \mathrm{NaHCO}_{3}\) solution (density of solution \(=\) \(1.19 \mathrm{~g} / \mathrm{mL}\)

Two liquids A and B have vapor pressures of \(76 \mathrm{mmHg}\) and \(132 \mathrm{mmHg},\) respectively, at \(25^{\circ} \mathrm{C}\) What is the total vapor pressure of the ideal solution made up of (a) 1.00 mole of \(\mathrm{A}\) and 1.00 mole of \(\mathrm{B}\), and (b) 2.00 moles of \(\mathrm{A}\) and 5.00 moles of \(\mathrm{B}\) ?

Calculate the molality of each of the following solutions: (a) \(14.3 \mathrm{~g}\) of sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) in \(676 \mathrm{~g}\) of water, (b) 7.20 moles of ethylene glycol \(\left(\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}_{2}\right)\) in \(3546 \mathrm{~g}\) of water.
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