A saturated solution was formed when 4.48×10−2 L of argon, at a pressure of 1.0 atm and temperature of 25 C, was dissolved in 1.0 L of water. Calculate the Henry's law constant for argon in mol/atm.
I got an answer of 1.4*10^-3, maybe my units are wrong, idk but if u could show your work that would be awesome
To calculate the Henry's law constant for argon in mol/atm, we can use the Henry's law equation:
C = k × P
where C is the concentration of the gas in mol/L, k is the Henry's law constant, and P is the partial pressure of the gas in atm.
In this case, we need to find the Henry's law constant (k) for argon. We are given that a saturated solution was formed when 4.48×10^(-2) L of argon, at a pressure of 1.0 atm and temperature of 25°C, was dissolved in 1.0 L of water.
First, we need to convert the volume of argon to concentration in mol/L:
C = (n_gas) / V_solution
where n_gas is the number of moles of argon and V_solution is the volume of the solution.
Using the ideal gas law equation:
PV = nRT
where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant (0.0821 L·atm/mol·K), and T is the temperature.
Let's calculate the number of moles (n_gas) of argon:
n_gas = PV / RT
= (1.0 atm) × (4.48×10^(-2) L) / ((0.0821 L·atm/mol·K) × (25 + 273.15 K))
≈ 6.00 × 10^(-4) mol
Next, let's calculate the concentration (C) of argon:
C = (6.00 × 10^(-4) mol) / (1.0 L)
= 6.00 × 10^(-4) mol/L
Now, we can use Henry's law equation to find the Henry's law constant (k):
C = k × P
Rearrange the equation to solve for k:
k = C / P
Substituting the given values:
k = (6.00 × 10^(-4) mol/L) / (1.0 atm)
= 6.00 × 10^(-4) mol/atm
Therefore, the Henry's law constant for argon is approximately 6.00 × 10^(-4) mol/atm.
To calculate the Henry's law constant for argon in mol/atm, we can use the equation:
C = k * P
Where:
C is the concentration of argon in mol/L (also known as solubility)
k is the Henry's law constant in mol/(L*atm)
P is the partial pressure of argon in atm
In the given problem, the concentration of argon is not provided directly, but we can calculate it using the provided information.
First, convert 4.48×10^(-2) L to moles of argon:
Volume of Ar = 4.48×10^(-2) L
Molar mass of Ar = 39.95 g/mol
Convert L to mol: (4.48×10^(-2) L) * (1 mol/22.4 L) = 2×10^(-3) mol Ar
Next, we calculate the concentration (C) of argon in mol/L:
Concentration (C) = (2×10^(-3) mol) / (1 L) = 2×10^(-3) mol/L
Now, we can rearrange the equation C = k * P to solve for the Henry's law constant (k):
k = C / P
Substituting the values we obtained:
k = (2×10^(-3) mol/L) / (1.0 atm)
Therefore, the Henry's law constant for argon is:
k = 2×10^(-3) mol/atm (or 2.0×10^(-3) mol/atm)
So, your answer of 1.4×10^(-3) was close, but you made an error in calculating the concentration. The correct Henry's law constant for argon is 2.0×10^(-3) mol/atm.
2.2x10^-2
Start with the Ideal Gas Law
PV = nRT
or
n = PV / RT
K = n/P (from Henry's Law)
n/P = V/RT (from the Ideal Gas Law)
or
K = V/RT (by combining the last two equations)
(V = liters, R = 0.0821 L.atm/K.mol, T = 298K)
Substitute to get the constant, K