At ordinary body temperature (37 °C), the solubility of N 2 in water at ordinary atmospheric pressure (1.0 atm) is 0.015 g/L. Air is approximately 78 mol % N 2 . (b) At a depth of 100 ft in water, the external pressure is 4.0 atm. What is the solubility of N 2 from air in blood at this pressure?

Hello everyone today. We are being given the following problem and asked us to solve for it. So it says the scalability of carbon dioxide and blood at standard atmospheric pressure is one atmospheres and at a temperature of 37 degrees Celsius is 1.0 to 6 g per liter. If the air approximately if the air is approximately 60.3% moles of C. O. To calculate the suitability of CO two from the air in blood at a depth of ft in water where the external pressure is 6.9 atmospheres. So this is a multi step problem. So the first step we want to do is we want to calculate our salt ability in malls per leader or polarity. And so by doing that we have to take our density of C. 02. So we have to take that 1.0 to 6 g per one liter of blood. And we have to multiply that by the molar mass of C. 02 itself. So we say that one mole sio two is equal to 44.1 g. Once our units can cancel out we are going to get 0. moles of our C. 02. All over one leader or just leaders. The second thing we wanna do is we want to calculate the partial. So we're gonna say we want to calculate the partial pressure at one atmospheres. And so by doing that we see that P. O. P. C. 02 are partial pressure of carbon dioxide is going to be equal to 0.3% mold Times one atmosphere to give us 0. atmospheres. The third thing we wanna do is we want to calculate so we're going to calculate for K. And so we have our cell viability of C. 02 is going to be equal to K. Is equal to the partial pressure of CO two. Rearranging this equation. We get that K. Is equal to our ability of C. 02 over our partial pressure of CO two. And when we plug these values in we're going to get 0.23 moles of C. 02. And then we're going to get all of that over that leaders. But then we have to take that and we have to put that over our partial pressure as indicated by the question as 0.3 atmospheres. To give us a total of 0.766 moles of C. 02 over Leaders times atmospheres. And our last and final step or one of our last our final step step four is going to be to calculate the partial pressure of C. 02 at 6.9 atmospheres. And so by doing that, we see that PCO two is equal to 0.03% times 6.9 atmospheres. A total of 0.207 atmospheres. And lastly we're going to calculate s Or our ability of CO2 At 6. atmospheres. And so we're gonna show that are selling ability of C. 02 is going to equal art zero point moles that we calculated earlier of C. 02 over Leaders Times Atmospheres. We're gonna multiply that by 0.027 atmospheres that we solved any question above. our units of atmospheres are going to cancel out and we are going to be left with a final answer of 0.159 moles of C. 02 over leaders which is also the malaria. T. I hope this helped. And until next time.

Ch.13 - Properties of Solutions

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Brown 14th Edition

Ch.13 - Properties of Solutions

Problem 107b

Chapter 13, Problem 107b

At ordinary body temperature (37 °C), the solubility of N₂ in water at ordinary atmospheric pressure (1.0 atm) is 0.015 g/L. Air is approximately 78 mol % N₂. (b) At a depth of 100 ft in water, the external pressure is 4.0 atm. What is the solubility of N₂ from air in blood at this pressure?

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Identify that the problem involves Henry's Law, which relates the solubility of a gas in a liquid to the partial pressure of the gas above the liquid.

Recall Henry's Law formula: \( S = k_H \cdot P \), where \( S \) is the solubility, \( k_H \) is Henry's Law constant, and \( P \) is the partial pressure of the gas.

Determine the partial pressure of \( N_2 \) at 4.0 atm. Since air is 78% \( N_2 \), calculate \( P_{N_2} = 0.78 \times 4.0 \text{ atm} \).

Use the given solubility of \( N_2 \) at 1.0 atm (0.015 g/L) to find \( k_H \) by rearranging Henry's Law: \( k_H = \frac{S}{P} \).

Calculate the new solubility \( S' \) at 4.0 atm using the previously determined \( k_H \) and the new partial pressure: \( S' = k_H \cdot P_{N_2} \).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Henry's Law

Henry's Law states that the amount of gas that dissolves in a liquid at a given temperature is directly proportional to the partial pressure of that gas above the liquid. This principle is crucial for understanding how gases like N2 dissolve in blood under varying pressures, as it allows us to predict solubility changes with pressure.

Partial Pressure

Partial pressure refers to the pressure exerted by a single component of a gas mixture. In the context of the question, the partial pressure of N2 in air can be calculated from its molar percentage and the total pressure. This value is essential for applying Henry's Law to determine the solubility of N2 in blood at increased depths.

Gas Solubility in Liquids

The solubility of gases in liquids is influenced by temperature, pressure, and the nature of the gas and solvent. In this scenario, understanding how N2 behaves in blood under different pressures helps in calculating its solubility at a depth of 100 ft, where the pressure is significantly higher than at the surface.

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A lithium salt used in lubricating grease has the formula LiC nH2n + 1O2. The salt is soluble in water to the extent of 0.036 g per 100 g of water at 25 °C. The osmotic pressure of this solution is found to be 57.1 torr. Assuming that molality and molarity in such a dilute solution are the same and that the lithium salt is completely dissociated in the solution, determine an appropriate value of n in the formula for the salt.

Fluorocarbons (compounds that contain both carbon and fluorine) were, until recently, used as refrigerants. The compounds listed in the following table are all gases at 25 °C, and their solubilities in water at 25 °C and 1 atm fluorocarbon pressure are given as mass percentages. (c) Infants born with severe respiratory problems are sometimes given liquid ventilation: They breathe a liquid that can dissolve more oxygen than air can hold. One of these liquids is a fluorinated compound, CF₃(CF₂)₇Br. The solubility of oxygen in this liquid is 66 mL O₂ per 100 mL liquid. In contrast, air is 21% oxygen by volume. Calculate the moles of O₂ present in an infant’s lungs (volume: 15 mL) if the infant takes a full breath of air compared to taking a full “breath” of a saturated solution of O₂ in the fluorinated liquid. Assume a pressure of 1 atm in the lungs.

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Textbook Question

A series of anions is shown below:

The anion on the far right is called 'BARF' by chemists, as its common abbreviation sounds similar to this word. (d) Tetrabutylammonium, (CH3CH2CH2CH2)4N + is a bulky cation. Which anion, when paired with the tetrabutylammonium cation, would lead to a salt that will be most soluble in nonpolar solvents?

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Textbook Question

A series of anions is shown below:

The anion on the far right is called 'BARF' by chemists, as its common abbreviation sounds similar to this word. (a) What is the central atom and the number of electronpair domains around the central atom in each of these anions?

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Textbook Question

A series of anions is shown below:

The anion on the far right is called 'BARF' by chemists, as its common abbreviation sounds similar to this word. (b) What is the electron-domain geometry around the central B in BARF?

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At ordinary body temperature (37 °C), the solubility of N2 in water at ordinary atmospheric pressure (1.0 atm) is 0.015 g/L. Air is approximately 78 mol % N2. (b) At a depth of 100 ft in water, the external pressure is 4.0 atm. What is the solubility of N2 from air in blood at this pressure?

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Key Concepts

Henry's Law

Partial Pressure

Gas Solubility in Liquids

At ordinary body temperature (37 °C), the solubility of N₂ in water at ordinary atmospheric pressure (1.0 atm) is 0.015 g/L. Air is approximately 78 mol % N₂. (b) At a depth of 100 ft in water, the external pressure is 4.0 atm. What is the solubility of N₂ from air in blood at this pressure?