The vapor pressure of pure water at 90celsius is normally 525.8 but decreased 483mmHg upon addition of an unknown amount of KNO3 to 286.9g H2O at this temperature. how many grams KNO3 were added. ASSUME COMPLETE DISSOCIATION OF SOLUTE. Tried all solutions provided and still got it wrong. Please help.
delta P = Xsolute*P<su>osolvent
483 = Xsolute*525.8
X = ??
KNO3 dissociates into two particles so take half of that = xx
Then moles KNO3/(total moles) = xx
moles H2O = 286.9/18.015 = yy
If we let Y = moles KNO3, then
[Y/(Y+moles H2O)] = xx and solve for Y, then Y x molar mass KNO3 = grams. I would check this out by changing g KNO3 to moles, then times 2 and add to moles H2O to find total moles. XKNO3 = ?? and plug into delta P = X*Po and see if you get 483.
To solve this problem, you need to use the concept of Raoult's law and Dalton's law of partial pressures. Here's the step-by-step solution:
Step 1: Understand the problem and identify the given information.
- The vapor pressure of pure water at 90°C is normally 525.8 mmHg.
- The vapor pressure decreased by 483 mmHg upon addition of KNO3.
- The weight of water (solvent) is given as 286.9 g.
Step 2: Apply Raoult's law to determine the moles of water present.
Raoult's law 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. Assuming complete dissociation of KNO3, the mole fraction of water (solvent) can be calculated as follows:
Molecular weight of H2O = 18 g/mol
Molar mass of KNO3 = 101.1 + 14 + (3 * 16) = 101.1 + 14 + 48 = 163.1 g/mol
Moles of H2O = (weight of H2O) / (molecular weight of H2O) = 286.9 g / 18 g/mol = 15.94 mol
Step 3: Determine the moles of KNO3 added.
Since KNO3 is assumed to completely dissociate in water, each mole of KNO3 yields one mole of K+ ions and one mole of NO3- ions. Thus, the total moles of KNO3 added can be calculated as follows:
Moles of KNO3 = 2 * (molar ratio of K+ or NO3- to H2O) * (moles of H2O)
The molar ratio of K+ or NO3- to H2O can be determined from the change in vapor pressure.
Change in vapor pressure = Vapor pressure of pure water - Vapor pressure of the solution
483 mmHg = 525.8 mmHg - Vapor pressure contributed by the K+ or NO3- ions
The vapor pressure contributed by the K+ or NO3- ions can be calculated using Dalton's law of partial pressures. Since both ions are assumed to be equal in number, we can divide the change in vapor pressure by 2 to obtain the vapor pressure contributed by each ion.
Vapor pressure contributed by K+ or NO3- = 483 mmHg / 2 = 241.5 mmHg
Now, we can calculate the molar ratio:
Molar ratio of K+ or NO3- to H2O = (vapor pressure contributed by K+ or NO3-) / (vapor pressure of pure water)
= 241.5 mmHg / 525.8 mmHg = 0.4599
Finally, calculate the moles of KNO3 added:
Moles of KNO3 = 2 * (molar ratio of K+ or NO3- to H2O) * (moles of H2O)
= 2 * 0.4599 * 15.94 mol = 14.65 mol
Step 4: Calculate the mass of KNO3 added.
Mass of KNO3 = (moles of KNO3) * (molar mass of KNO3)
= 14.65 mol * 163.1 g/mol = 2,391.02 g
Therefore, approximately 2391.02 grams of KNO3 were added to the solution.
Make sure to double-check your calculations and units to ensure accuracy.