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Grade: 12
The vapour pressure of water is 12.3 kpa at 300k. calculate the vapour pressure of 1 molal solution of a non volatile  solute in it?
2 years ago

Answers : (1)

23506 Points
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
        ... (2)
Here we know that number of moles of solute = 1 moles
And need to find the moles of solvent (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
2 years ago
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