Most fish need at least 4.00 ppm dissolved O2 for survival.
(a) What is this concentration in M (assume the density of the
solution is 1.00 g/mL)?
(b) What partial pressure of O2 above the water is needed to
obtain this concentration at 10. °C? The Henry’s law
constant for O2 at this temperature is 1.71 ×10-3 mol/L-atm
I'm not really sure how to even start :/
a.
4 ppm = 4 mg/L.
mols O2 = grams/molar mass and since that many mols are in 1 L that is the M.
b.
concn(M) = K*P
You know M from part a and you K from the problem, solve for partial pressure of the gas.
Thank you so much for your help!
To solve this problem, we need to convert the given concentration of dissolved oxygen from ppm (parts per million) to molarity (M) and then calculate the partial pressure of oxygen using Henry's law.
(a) To convert ppm to Molarity (M), we need to know the molar mass of oxygen and the density of the solution.
The molar mass of oxygen (O₂) is 32.00 g/mol.
Let's assume that 1 L of the solution weighs 1.00 kg (since the density is given as 1.00 g/mL = 1.00 kg/L).
To convert ppm to M, we use the formula:
ppm = (mass of solute / mass of solution) × 10⁶
In this case, the mass of solute (dissolved oxygen) is given by:
mass of solute = (ppm / 10⁶) × mass of solution
mass of solute = (4.0 ppm / 10⁶) × 1.00 kg = 4.00 × 10⁻⁶ kg
Now, convert the mass of solute (in kg) to moles (mol):
moles of solute = mass of solute / molar mass
moles of solute = (4.00 × 10⁻⁶ kg) / (32.00 g/mol) = 1.25 × 10⁻⁷ mol
Finally, calculate the molarity (M) using the formula:
Molarity (M) = moles of solute / volume of solution (in L)
In this case, since the density is given as 1.00 g/mL = 1.00 kg/L, the volume of the solution is equal to the mass of the solution (1.00 kg):
Molarity (M) = (1.25 × 10⁻⁷ mol) / 1.00 L = 1.25 × 10⁻⁷ M
Therefore, the concentration of dissolved oxygen in Molarity (M) is 1.25 × 10⁻⁷ M.
(b) To calculate the partial pressure of oxygen above the water using Henry's law, we will use the formula:
Partial pressure of oxygen = Henry's law constant × Molarity of oxygen
Given:
Henry's law constant (k) = 1.71 × 10⁻³ mol/L-atm
Molarity of oxygen (M) = 1.25 × 10⁻⁷ M
Partial pressure of oxygen = (1.71 × 10⁻³ mol/L-atm) × (1.25 × 10⁻⁷ M)
Partial pressure of oxygen = 2.14 × 10⁻¹⁰ atm
Therefore, the partial pressure of oxygen required to obtain a concentration of 4.00 ppm dissolved oxygen is 2.14 × 10⁻¹⁰ atm.
No worries! Let's break it down step by step.
(a) To convert parts per million (ppm) to molar concentration (M), we need to know the molecular weight of oxygen (O2). The molecular weight of O2 is 32.0 g/mol (16.0 g/mol for each oxygen atom).
Given that the density of the solution is 1.00 g/mL, we can assume that 1 L of solution has a mass of 1 kg (1000 g). Since we are dealing with ppm, we can say that 4.00 ppm means 4.00 mg of O2 per kg of solution.
To convert mg to moles, we need to divide by the molecular weight of O2:
4.00 mg / 32.0 g/mol = 0.125 mol.
Now, we can calculate the molar concentration (M) in mol/L by dividing the number of moles by the volume of the solution in liters (1 L in this case):
0.125 mol / 1 L = 0.125 M.
So, the concentration of O2 in molar units (M) is 0.125 M.
(b) Henry's Law relates the concentration of a gas in a liquid to its partial pressure above the liquid. The equation is given as:
C = k * P
where C is the concentration of the gas in mol/L, k is Henry's Law constant, and P is the partial pressure of the gas in atm.
Given that the concentration (C) is 0.125 M and the Henry's Law constant (k) is 1.71 × 10^(-3) mol/L-atm, we can rearrange the equation to solve for P:
P = C / k.
Substituting the values:
P = 0.125 M / (1.71 × 10^(-3) mol/L-atm).
Simplifying:
P = 73.1 atm.
So, the partial pressure of O2 above the water needed to obtain a concentration of 4.00 ppm (0.125 M) at 10. °C is approximately 73.1 atm.