The vapour pressure of water is 12.3 kPa at 300 K. Calculate the

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 Multiple Choice QuestionsShort Answer Type

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222.

An aqueous solution of 2 percent non-volatile solute exerts a pressure of 0.096 atm at the boiling poine of the solvent. What is the molecular mass of the solute ?

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223. The vapour pressure of water is 12.3 kPa at 300 K. Calculate the vapour pressure of a one molal solution of a non-volatile, non-ionic solute in water.


Solution
Step-1 what is given in problem
Vapour pressure of pure water = 12.3 kPa
We have 1 molal solution that means we have 1 moles of solute in 1kg of solvent
Step – 2 Find vapour pressure
The formula of vapour pressure when non-volatile liquid is added
Pressure = vapour pressure of pure liquid × molar fraction of liquid

P=P(pure)X  ...........1                                          

 Here we know the value of P(pure)so need to find the value of molar fraction X

moles fraction of compound = number of moles of compoundstotal number of moles      .....2
  
Here we know that number of moles of solute =1 moles
And need to find the moles of solvent (water)
                             
number of moles of water= mass of watermolar mass of water 

And we have mass = 1 kg and 1 kg is equal to 1000g
And molar mass of water H2O = 2×1 + 16 = 18 g/mol
Plug the value in equation (3) we get
Number of moles of water =1000 g / (18g/mol  )  =55.56 moles
Plug the value in equation (2) we get
The formula of molar fraction =55.5/(55.5+1)  = 0.9823
Now plug in equation (1 )we get
Partial fraction = 12.3×0.9823 = 12.08
 
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