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A solution is prepared by dissolving 80.0 g of sucrose, C12H22O11, in 250. g of water...

A solution is prepared by dissolving 80.0 g of sucrose, C12H22O11, in 250. g of water at 25 °C. What is the vapour pressure of the solution if the vapour pressure of water at 25 °C is 23.76 mmHg?

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Answer #1

Raoults law

P0 - Ps/Po = Xb

no of mole of C12H22O11 = w/mwt = 80/342 = 0.234 mole

no of mole of Water = w/mwt = 250/18 = 13.89 mole

molefraction of solute(XC12H22O11) = nC12H22O11/(nC12H22O11+nH2O)

                            = 0.234/(13.89+0.234) = 0.0166


P0 = vapor pressure of solvent = 23.76 mmHg

Ps = vapor pressure of solution = x


((23.76-x)/23.6) = 0.0166

Ps = 23.37 mmhg

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