The Henry's Law constant for O2 in water is 1.28 x 10 ^ -3 M/atm. How many grams of O2 will dissolve in a 12 L fish tank on a day when atm is .977 and air is 18% O2?

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I looked on the net at this site to see the formula you must be using for Henry's Law (many different versions in use). You can read more about it here.

http://en.wikipedia.org/wiki/Henry%27s_law

You must be using c = pk; therefore,
c = (0.18*0.977)*1.28E-3
c = 2.25E-4 M. Convert that to grams (g = moles x molar mass) for the mass O2 in 1L, convert that to 12L.

To calculate the number of grams of O2 that will dissolve in the fish tank, we can use Henry's Law, which states that the amount of gas dissolved in a liquid is directly proportional to the partial pressure of the gas.

Step 1: Convert the given partial pressure of O2 from atm to units of Molarity (M).
Henry's Law constant (k) = 1.28 x 10^-3 M/atm
Partial pressure of O2 (P) = 0.977 atm

To convert atm to Molarity, we need to use the ideal gas law equation:
PV = nRT

Rewriting the ideal gas law to solve for molarity:
M = n/V

Where:
M = Molarity in moles/Liter
n = Number of moles
V = Volume in Liters
R = Ideal gas constant (0.0821 L·atm/(mol·K))
T = Temperature in Kelvin (assumed constant)

Given that the atmospheric pressure and temperature are constant, we can ignore the temperature term for this calculation.

M = P/R

M = 0.977 atm / 0.0821 L·atm/(mol·K)

M ≈ 11.9 M

Step 2: Calculate the moles of O2 that will dissolve using Henry's Law.
According to Henry's Law, we know that the concentration of O2 (C) will be equal to k (Henry's Law constant) multiplied by the partial pressure of O2 (P).

C = k * P

C = (1.28 x 10^-3 M/atm) * (11.9 M)

C ≈ 1.5232 x 10^-2 M

Step 3: Convert moles of O2 to grams of O2.
To convert moles to grams, we need to use the molar mass of O2, which is 32 g/mol.

Grams of O2 = moles of O2 * molar mass of O2

Grams of O2 = (1.5232 x 10^-2 M) * (32 g/mol)

Grams of O2 ≈ 0.487 g

Therefore, approximately 0.487 grams of O2 will dissolve in the 12 L fish tank on a day when the atmospheric pressure is 0.977 atm, and the air contains 18% O2.