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Aleks [24]
2 years ago
11

The solubility of N2 in blood can be a serious problem (the "bends") for divers breathing compressed air (78% N2 by volume) at d

epths greater than 50 ft. Find the volume (in mL) of N2, measured at 25°C and 1.00 atm, released per liter of blood when a diver at a depth 50. ft rises to the surface. The total pressure (water weight plus atmospheric pressure) on the diver at 50. ft is 2.47535 atm. (kH for N2 in water at 25°C is 7.0 × 10−4 mol/L·atm.)
Chemistry
1 answer:
OLga [1]2 years ago
5 0

Answer:

The volume is 19.7 mL

Explanation:

<u>Step 1</u>: Given data

Pressure at sea level = 1.00 atm

Pressure at 50 ft = 2.47535 atm

kH for N2 in water at 25°C is 7.0 × 10−4 mol/L·atm

Molarity (M) = kH x P

<u>Step 2</u>: Calculate molarity

M at sea level:

M = 7.0*10^-4 * (1.00atm * 0.78) = 5.46*10^-4 mol/L

M at 50ft:

M = 7.0*10^-4 * (2.47535atm * 0.78) = 13.5*10^-4 mol/L

We should find the volume of N2. To find the volume whe have to find the number of moles first. This we calculate by calculating the difference between M at 50 ft and M at sea level.

13.5*10^-4 mol/L - 5.46*10^-4 mol/L = 8.04*10^-4 mol/L

Step 3: Calculate volume

P*V=nRT

with P = 1.00 atm

with V = TO BE DETERMINED

with n =  8.04*10^-4 mol/L  *1L = 8.04*10^-4

with R= 0.0821 atm * L/ mol *K

with T = 25 °C = 273+25 = 298 Kelvin

To find the volume, we re-organize the formula to: V=nRT/P

V= (8.04*10^-4 mol * 0.0821 (atm*L)/(mol*K)* 298K ) / 1.00atm = 0.0197L = 19.7ml

The volume is 19.7 mL

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Norfenefrine (C₈H₁₁NO₂).

<h3>Further explanation</h3>

We will solve a case related to one of the colligative properties, namely freezing point depression.

The freezing point of the solution is the temperature at which the solution begins to freeze. The difference between the freezing point of the solvent and the freezing point of the solution is called freezing point depression.

\boxed{ \ \Delta T_f = T_f(solvent) - T_f(solution) \ } \rightarrow \boxed{ \ \Delta T_f = K_f \times molality \ }

<u>Given:</u>

A mysterious white powder could be,

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When 82 mg of the powder is dissolved in 1.50 mL of ethanol (density = 0.789 g/cm³, normal freezing point −114.6°C, Kf = 1.99°C/m), the freezing point is lowered to −115.5°C.

<u>Question: </u>What is the identity of the white powder?

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Let us identify the solute, the solvent, initial, and final temperatures.

  • The solute = the powder
  • The solvent = ethanol
  • The freezing point of the solvent = −114.6°C
  • The freezing point of the solution = −115.5°C

Prepare masses of solutes and solvents.

  • Mass of solute = 82 mg = 0.082 g
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We must prepare the solvent mass unit in kg because the unit of molality is the mole of the solute divided by the mass of the solvent in kg.

The molality formula is as follows:

\boxed{ \ m = \frac{moles \ of \ solute}{kg \ of \ solvent} \ } \rightarrow \boxed{ \ m = \frac{mass \ of \ solute \ (g)}{molar \ mass \ of \ solute \times kg \ of \ solvent} \ }

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\boxed{ \ \Delta T_f =  K_f \times \frac{mass \ of \ solute \ (g)}{molar \ mass \ of \ solute \times kg \ of \ solvent} \ }

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\boxed{ \ 0.9 = \frac{1.99 \times 0.082}{Mr \times 0.00118} \ }

\boxed{ \ Mr = \frac{0.16318}{0.9 \times 0.00118} \ }

We get \boxed{ \ Mr = 153.65 \ }

These results are very close to the molar mass of norfenefrine which is 153.18 g/mol. Thus the white powder is norfenefrine.

<h3>Learn more</h3>
  1. The molality and mole fraction of water brainly.com/question/10861444
  2. About the mass and density of ethylene glycol as an  antifreeze brainly.com/question/4053884
  3. About the solution as a homogeneous mixture  brainly.com/question/637791

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