12. The vapor pressure of water at 90°C is 0.692 atm. What is the vapor pressure (in atm) of a solution made by dissolving 3.68 mole(s) of CsF(s) in 1.00 kg of water? Assume that Raoult's law applies.

Answer : The vapor pressure (in atm) of a solution is, 0.679 atm

Explanation : Given,

Mass of = 1.00 kg = 1000 g

Moles of = 3.68 mole

Molar mass of = 18 g/mole

Vapor pressure of water = 0.692 atm

First we have to calculate the moles of .

Now we have to calculate the mole fraction of

Now we have to partial pressure of solution.

According to the Raoult's law,

where,

= vapor pressure of solution

= vapor pressure of water = 0.692 atm

= mole fraction of water = 0.938

Therefore, the vapor pressure (in atm) of a solution is, 0.679 atm

answer : third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol

explanation :

let us write down the given equations

equation 1 :

equation 2 :

equation 3 :

let us use hess's law to find out which 2 equations can be added to get the third equation.

let us reverse equation 2 and multiply it by 2. the new equation that we get is,

equation 4 :

let us add equation 1 & 4

the above equation is similar to equation 3 but has different value for dhrxn.

therefore third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol