the vapour pressure of a dillutes aqueous solution of glucose is 750 mm of hg at 373k. calculate mole fraction of solutes
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Ans- We know from Rault's law,
p0 - pr / p0 = n2 / n1 + n2 = mole fraction of solute
Here, n1 = no. of moles of solvent water = 1000/ 18 = 55.55 mol. p0 = 760 mm and pr = 750 mm
So, putting the value ,
mole fraction of solute = 760-750/760 = 0.013
Now molality of solution = no of moles of solute/ in 1kg water.
So, since55.55 moles of water contains 1Kg water so,
molality of Solute = 0.013 x 55.55 = 0.73
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