Question

The vapor pressure of water at 20 C is 17.5mmHg. What is the vapor pressure of water over a solution prepared from 200g of sucrose (`C_12H_22O_11`) and 112.3 g of water?

Answer 1

We use `"Raoult's Law..........."` and calculate a solution vapour pressure of `16*mm*Hg`.

Explanation:

`"Raoult's Law"` tells us that in an ideal solution, the vapour pressure of a component is proportional to the mole fraction of that component in solution.

`"Mole fraction of sucrose:"` `="Moles of sucrose"/"Moles of sucrose + moles of water"`

`chi_"sucrose"=((200*g)/(342.13*g*mol^-1))/((200*g)/(342.13*g*mol^-1)+(112.3*g)/(18.01*g*mol^-1))=0.0857`

`"Mole fraction of water:"`

`chi_"water"=((112.3*g)/(18.01*g*mol^-1))/((200*g)/(342.13*g*mol^-1)+(112.3*g)/(18.01*g*mol^-1))=0.914`

Note that by definition, `chi_"sucrose"+chi_"water"=1`

Because sucrose is involatile, the vapour pressure of the solution is proportional to the mole fraction of water:

`P_"solution"=chi_"water"xx17.5*mm*Hg`

`=0.914xx17.5*mm*Hg=16*mm*Hg`

We might have got better results if we used ethyl alcohol or acetone as the solvent rather than water, given that these organic solvents would have expressed a REDUCED mole fraction with respect to water, and thus a GREATER diminution in vapour pressure with respect to that of the pure solvent....