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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 fresh water 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 pipe must be immersed approximately 336.1 meters deep into the sea.

Step by step solution

01

Understanding Osmotic Pressure

Osmotic pressure in a solution like seawater can be described by the formula \( \pi = i \cdot C \cdot R \cdot T \), where \( i \) is the van't Hoff factor, \( C \) is the molarity of the solution, \( R \) is the ideal gas constant \(0.0821 \frac{L \cdot atm}{K \cdot mol} \), and \( T \) is the temperature in Kelvin. Since NaCl dissociates into two ions (Na+ and Cl-), the van't Hoff factor \( i = 2 \). The temperature in Kelvin is \( T = 293K \) (since \( 20^{\circ}C + 273 = 293K \)).
02

Calculate Osmotic Pressure

We now calculate the osmotic pressure using \( \pi = 2 \times 0.70 \times 0.0821 \times 293 \). This calculation gives \( \pi \approx 33.7 \ atm \). To proceed, we convert this pressure from atmospheres to pascals (Pa), knowing that \( 1 atm = 101325 Pa \). Therefore, \( \pi \approx 33.7 \times 101325 = 3,416,977.5 \ Pa \).
03

Calculating the Depth Needed

The osmotic pressure is equal to the hydrostatic pressure at the depth of immersion required to allow fresh water to permeate. This is given by \( P = \rho \cdot g \cdot h \), where \( \rho = 1030 kg/m^3 \) is the density of seawater, \( g = 9.81 m/s^2 \) is gravitational acceleration, and \( h \) is the depth in meters. Thus, rearrange the equation to find \( h = \frac{\pi}{\rho \cdot g} \). Plugging the values in gives \( h = \frac{3416977.5}{1030 \cdot 9.81} \approx 336.1 \ m \).

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Key Concepts

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

Van't Hoff Factor
When studying solutions, especially electrolytes like NaCl, the Van't Hoff factor (\( i \) ) is crucial for understanding how molecules behave when they dissolve. This factor represents the number of particles a compound dissociates into in solution. For example:
  • NaCl dissociates into two ions in a solution: Na+ and Cl-, so the Van't Hoff factor for NaCl is 2.
  • Non-electrolytes, like glucose, do not dissociate in solution, so their Van't Hoff factor is 1.
The Van't Hoff factor is also integral in calculating osmotic pressure. It amplifies the concentration of solute affecting the osmotic potential, which is why understanding this factor is crucial when predicting how solutions will interact, like seawater in our given problem. Remember: more ions lead to higher osmotic pressures, which is vital for calculating the necessary immersion depth of the pipe.
Molarity Calculation
Molarity (denoted as \( C \)), is a measure of the concentration of a solute in a solution. It is expressed as the number of moles of solute per liter of solution. In our exercise, calculating molarity is straightforward:
  • The given solution is 0.70 M NaCl, meaning 0.70 moles of NaCl are present per liter of solution.
  • Molarity is essential in determining osmotic pressure because it directly influences the calculated pressure via the formula \( \pi = i \cdot C \cdot R \cdot T \). As such, its accurate calculation is imperative for further steps in solving the problem.
Understanding molarity helps visualize how solute concentration impacts the behavior of a solution, vital for comprehending phenomena such as osmosis especially in biological and chemical processes.
Hydrostatic Pressure
Hydrostatic pressure refers to the pressure exerted by a fluid at equilibrium due to the force of gravity. It's calculated using the formula \( P = \rho \cdot g \cdot h \), where:
  • \( \rho \) is the fluid's density (for seawater, that's \(1030 \ kg/m^3\)).
  • \( g \) represents gravitational acceleration, approximately \(9.81 \ m/s^2\).
  • \( h \) is the height or depth of the fluid column.
In practical applications, like determining the immersion depth of a pipe, you match osmotic pressure with hydrostatic pressure. Hence, by re-arranging the hydrostatic pressure equation, one can solve for depth \( h \) once the pressure needed is known. This relationship helps explain freshwater movement through a membrane due to pressure differences, a principle widely used in understanding natural processes and industrial applications such as reverse osmosis.

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Most popular questions from this chapter

A quantity of \(7.480 \mathrm{~g}\) of an organic compound is dissolved in water to make \(300.0 \mathrm{~mL}\) of solution. The solution has an osmotic pressure of \(1.43 \mathrm{~atm}\) at \(27^{\circ} \mathrm{C}\). The analysis of this compound shows that it contains 41.8 percent \(\mathrm{C}, 4.7\) percent \(\mathrm{H}, 37.3\) percent \(\mathrm{O},\) and 16.3 percent \(\mathrm{N}\). Calculate the molecular formula of the compound.

A man bought a goldfish in a pet shop. Upon returning home, he put the goldfish in a bowl of recently boiled water that had been cooled quickly. A few minutes later the fish was found dead. Explain what happened to the fish.

The antibiotic gramicidin A can transport \(\mathrm{Na}^{+}\) ions into a certain cell at the rate of \(5.0 \times 10^{7} \mathrm{Na}^{+}\) ions/channel s. Calculate the time in seconds to transport enough \(\mathrm{Na}^{+}\) ions to increase its concentration by \(8.0 \times 10^{-3} M\) in a cell whose intracellular volume is \(2.0 \times 10^{-10} \mathrm{~mL}\).

Pheromones are compounds secreted by the females of many insect species to attract males. One of these compounds contains 80.78 percent \(\mathrm{C}, 13.56\) percent \(\mathrm{H},\) and 5.66 percent \(\mathrm{O}\). A solution of \(1.00 \mathrm{~g}\) of this pheromone in \(8.50 \mathrm{~g}\) of benzene freezes at \(3.37^{\circ} \mathrm{C}\). What are the molecular formula and molar mass of the compound? (The normal freezing point of pure benzene is \(\left.5.50^{\circ} \mathrm{C} .\right)\)

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