If the vapour pressure of
of water equals 0.03 atm (at 25°c), what is the vapour pressure of the solution which is produced from adding 5 mol of ethylene glycol to 1 l of water
at the same temperature
a 0.04 atm
b 0.027 atm
c 0.008 atm d 0.01 atm

b 0.027 atm

Explanation:

It is possible to solve this question using Raoult's law that says that the partial pressure of each component of an ideal mixture of liquids is equal to the vapour pressure of the pure component multiplied by its mole fraction in the mixture.

To know mole fraction of water you need to know how many moles of water have 1L. Moles are:

1L × (1000g/1L) × (1mol/18,01g) = 55,5 moles of water.

Mole fraction is:

55,5 moles of water / (55,5+5) total moles = 0,917

Thus, vapour pressure of the solution is:

0,917 × 0,03atm = 0,027atm

Right answer is:

b 0.027 atm

I hope it helps!

the answers would be

the first one and the second one