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.

Question

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.

DrBob222 · Answer

delta P = Xsolute*Posolvent 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.