When a specific amount of propanol (C3H8O) is added to 110.0 g of pure water at 60°C, the vapor pressure of water over the solution is lowered by 1.596 kPa. Given the vapor pressure of water at 60°C is 19.932 kPa, what is the mass of propanol added?
To find the mass of propanol added, we can utilize Raoult's Law, which states that the vapor pressure of a solution is equal to the mole fraction of the solvent multiplied by the vapor pressure of the pure solvent.
Let's first calculate the mole fraction of propanol (C3H8O) in the solution:
Molar mass of propanol (C3H8O) = 60.10 g/mol
Molar mass of water = 18.02 g/mol
Step 1: Calculate the mole fraction of water (xw):
Mole fraction of water = moles of water / total moles in the solution
To find the moles of water, we need to convert the mass of water to moles:
Moles of water = mass of water / molar mass of water
Moles of water = 110.0 g / 18.02 g/mol = 6.106 mol
Total moles in the solution = moles of water + moles of propanol
Since we are given that the mass of propanol is specific but unknown, let's represent the mass of propanol as "mp".
So, the total moles in the solution = 6.106 mol + (mp / 60.10 g/mol)
Now we can calculate the mole fraction of water (xw):
xw = 6.106 mol / (6.106 mol + (mp / 60.10 g/mol) )
Next, we need to use Raoult's Law to find the vapor pressure of the solution (P) and compare it to the vapor pressure of water (Po):
P = xw * Po
Given that the vapor pressure of water (Po) is 19.932 kPa and the vapor pressure of the solution (P) is reduced by 1.596 kPa, we have:
P = 19.932 kPa - 1.596 kPa = 18.336 kPa
Finally, we can substitute the values into Raoult's Law to calculate the mole fraction of water:
18.336 kPa = xw * 19.932 kPa
xw = 18.336 kPa / 19.932 kPa
Now we can substitute the calculated mole fraction of water (xw) back into the equation for the mole fraction of water:
18.336 kPa / 19.932 kPa = 6.106 mol / (6.106 mol + (mp / 60.10 g/mol) )
Simplifying the equation, we get:
18.336 / 19.932 = 6.106 / (6.106 + mp / 60.10)
Now, solve for mp (mass of propanol):
Divide both sides of the equation by (6.106 + mp / 60.10):
(18.336 / 19.932) = 6.106 / (6.106 + mp / 60.10)
Cross multiply:
(18.336 / 19.932) * (6.106 + mp / 60.10) = 6.106
Multiply both sides by (6.106 + mp / 60.10):
(18.336 / 19.932) * 6.106 + (18.336 / 19.932) * (mp / 60.10) = 6.106
Distribute the first term on the left side:
(18.336 / 19.932) * 6.106 + (mp / 60.10) * (18.336 / 19.932) = 6.106
Multiply both sides by 60.10 to get rid of the fraction:
((18.336 / 19.932) * 6.106 * 60.10) + mp * (18.336 / 19.932) = 6.106 * 60.10
Calculate the left side:
(mp * (18.336 / 19.932)) = ((6.106 * 60.10) - ((18.336 / 19.932) * 6.106 * 60.10))
Now, isolate mp by dividing both sides by (18.336 / 19.932):
mp = ((6.106 * 60.10) - ((18.336 / 19.932) * 6.106 * 60.10)) / (18.336 / 19.932)
Evaluating this expression gives us the mass of propanol added to the solution.
Psoln = Xsolvent*Po pure solvent.
You have Po H2O. Psoln = Po-1.596. Substitute and solve for Xsolvent.
Then XH2O = [nH2O/(nH2O + nPropanol)]
Substitute and solve for nPropanol and convert that to grams. Post your work if you gst stuck.