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_11C12H22O11) and 112.3 g of water?

1 Answer
anor277
Dec 28, 2016

We use "Raoult's Law..........."Raoult's Law........... and calculate a solution vapour pressure of 16*mm*Hg16⋅mm⋅Hg.

Explanation:

"Raoult's Law"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:"Mole fraction of sucrose: ="Moles of sucrose"/"Moles of sucrose + moles of water"=Moles of sucroseMoles 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))=0.0857χsucrose=200⋅g342.13⋅g⋅mol−1200⋅g342.13⋅g⋅mol−1+112.3⋅g18.01⋅g⋅mol−1=0.0857

"Mole fraction of water:"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))=0.914χwater=112.3⋅g18.01⋅g⋅mol−1200⋅g342.13⋅g⋅mol−1+112.3⋅g18.01⋅g⋅mol−1=0.914

Note that by definition, chi_"sucrose"+chi_"water"=1χsucrose+χwater=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*HgPsolution=χwater×17.5⋅mm⋅Hg

=0.914xx17.5*mm*Hg=16*mm*Hg=0.914×17.5⋅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....

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