Question

What is the vapor pressure of a solution of 300 g of ethylene glycol, HOCH₂CH₂OH (mol wt = 62.1) and 200 g of water (a 60% solution by weight) at 100 ºC. The vapor pressure of pure water is 760 torr at 100 ºC (the boiling point of water).

Answer 1

Answer –

Given, mass of solution = 300 g , mass of water = 200 g that is 60% solution by weight

Mass of ethylene glycol =300 g

First calculate the moles of each

Moles of ethylene glycol = 300 g / 62.1 g.mol^-1

                                         = 4.83 moles

Moles of water = 200 g / 18.015 g.mol^-1

                         = 11.10 moles

Now calculate the moles fraction of water

Total moles = 4.83+11.10 = 15.933 moles

Moles fraction of wter = 11.10 /15.933 moles

                                     = 0.697

Now need to use the Raoult’s law

P(H₂O) = X(H₂O) * P^o(H₂O)

The P^o(H₂O) is the vapor pressure of pure water and it is 760 torr

So, P(H₂O) = 0.697 * 760 torr

                   = 529.6 torr

The vapor pressure of a solution is 529.6 torr