The vapour pressure of water is 12.3 kPa at 300 K. Calculate vapour pressure of 1 molal solution of a non-volatile solute in it.

The vapour pressure of water is 12.3 kPa at 300 K. Calculate vapour pressure of 1 molal solution of a non-volatile solute in it.

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

         ... (2)

Here we know that number of moles of solute = 1 moles

And need to find the moles of solvent (water)

                              ...(3)

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