How many grams of O2 will dissolve in 2.50L of H2O that is constant with pure O2 at 1.00atm?
To answer this question, we need to apply the concept of Henry's Law. Henry's Law states that the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid.
The equation for Henry's Law is:
C = k * P
Where:
C is the concentration of the dissolved gas in the liquid (in mol/L or Molarity).
k is Henry's Law constant, specific for each gas and solvent.
P is the partial pressure of the gas above the liquid (in atm).
Since we are given the constant for pure O2 at 1.00 atm, we can use this information to find the value of k for O2 in H2O. However, the question asks for the mass of O2 that will dissolve, so we need to convert the concentration into grams.
To convert from concentration (C) to grams, we can use the following equation:
Mass = Concentration * Volume * Molar Mass
Let's go step by step to solve the problem:
Step 1: Determine the value of k for O2 in H2O.
Unfortunately, we don't have the value of k for O2 in H2O. However, we can assume the value of k to be approximately equal to the solubility of O2 in water at standard temperature and pressure (STP), which is 0.036 mol/L/atm.
Step 2: Calculate the concentration of O2 in H2O.
Since we know the partial pressure of O2 at 1.00 atm, we can use Henry's Law equation to calculate the concentration of O2 in H2O:
C = k * P
C = 0.036 mol/L/atm * 1.00 atm
C = 0.036 mol/L
Step 3: Convert concentration to grams.
Now, we need to convert the concentration of O2 from moles per liter to grams. To do this, we need to know the molar mass of O2, which is approximately 32 g/mol.
Mass = Concentration * Volume * Molar Mass
Mass = 0.036 mol/L * 2.50 L * 32 g/mol
Mass = 2.88 grams
Therefore, approximately 2.88 grams of O2 will dissolve in 2.50 liters of H2O at 1.00 atm.
To calculate the amount of O2 that will dissolve in water, 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 above the liquid.
The equation for Henry's Law is:
C = k * P
Where:
C is the concentration of the dissolved gas
k is the Henry's Law constant for the specific gas and solvent
P is the partial pressure of the gas
First, we need to find the value of the Henry's Law constant (k) for O2 in water. The value of k depends on temperature, so we will assume a standard temperature of 25°C.
The Henry's Law constant for O2 in water at 25°C is approximately 1.3 x 10^-3 M/atm.
Next, we need to convert the volume of water (2.50L) into units of moles. Since we are given the pressure in atm, we will use the ideal gas law to convert the volume to moles:
PV = nRT
Where:
P is the pressure (1.00 atm)
V is the volume (2.50L)
n is the number of moles
R is the ideal gas constant (0.0821 L.atm/mol.K)
T is the temperature (we will assume 25°C, which is 298K)
Solving for n, we get:
n = (P * V) / (R * T)
n = (1.00 atm * 2.50L) / (0.0821 L.atm/mol.K * 298K)
n = 0.1058 moles
Now, we can calculate the concentration (C) of O2 in water using Henry's Law:
C = k * P
C = (1.3 x 10^-3 M/atm) * (1.00 atm)
C = 1.3 x 10^-3 M
To find the mass of O2 that will dissolve in the water, we need to multiply the concentration (C) by the molar mass of O2 (32g/mol) and by the volume of water:
mass = C * molar mass * volume
mass = (1.3 x 10^-3 M) * (32g/mol) * (2.50L)
mass = 0.052g
Therefore, approximately 0.052 grams of O2 will dissolve in 2.50 liters of water at a constant pressure of 1.00 atm.