Make a solutions by dissolving 4 lbs of sulfur dioxide in 100 lbs of water then heat to 30 degress C. what is the partial pressure of the sulfur dioxide over the solution specified?
300 mm Hg
To determine the partial pressure of sulfur dioxide (SO2) over the solution, we need to use the concept of Henry's Law, which states that the partial pressure of a gas above a liquid is proportional to the concentration of the gas in the liquid.
First, we need to calculate the concentration of sulfur dioxide in the solution. We can use the formula:
Concentration (in moles per liter) = mass of solute / molar mass
The molar mass of sulfur dioxide (SO2) is approximately 64.06 g/mol.
Given that we are dissolving 4 lbs of sulfur dioxide in 100 lbs of water, we first need to convert the mass of sulfur dioxide to grams:
4 lbs = 4 lbs * 453.592 g/lb = 1814.366 g
Now, we can calculate the concentration of sulfur dioxide:
Concentration (in moles per liter) = 1814.366 g / 64.06 g/mol = 28.35 mol/L
Next, we need to convert the concentration to mol/kg since the given water mass is in kg:
Concentration (in mol/kg) = 28.35 mol/L * (1000 g/1 kg) = 28.35 mol/kg
Now, we can apply Henry's Law, which states that the partial pressure is proportional to the concentration:
Partial Pressure (in atm) = Henry's Law constant * Concentration
The Henry's Law constant for sulfur dioxide is approximately 1630 atm/mol/kg.
Partial Pressure (in atm) = 1630 atm/mol/kg * 28.35 mol/kg
Multiplying the values:
Partial Pressure = 46310.5 atm
Therefore, the partial pressure of sulfur dioxide over the specified solution is approximately 46310.5 atm when heated to 30 degrees Celsius.