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Title: A solution is prepared by dissolving 70.0 g of sucrose, C12H22O11, in 250. g of water at 25 °C. ... Post by: megan213r on Dec 8, 2016 A solution is prepared by dissolving 70.0 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?
Title: Re: A solution is prepared by dissolving 70.0 g of sucrose, C12H22O11, in 250. g of water at 25 °C. ... Post by: psyche360 on Dec 9, 2016 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 |