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Lowering of vapour pressure due to solute in 1 molal aqueous solution at 100°C is BOOK ANSWER - 13.44 Torr MY SOLUTION Since molality is 1 mole kg-1 That's mean 1 mole of solute is present in 1 kg of solvent (H2O) No. Of moles of Water = 1000g/18 = 55.55 No. Of moles of solute = 1 Mole Fraction of solute = 1/1+55.55 = 0.0176 .... (I) Pressure of solvent (assuming ideal behaviour) PV = nRT P = 55.55 x 0.0821 x 373 / 1 (d of water = 1) P = 1701.2774 atm 1 atm = 760 Torr psolvent = 1292970.8 Torr We know from Raoult Law ∆p = Xsolute . P(solvent) ∆p = 0.0176 x 1292970.8 ∆p = 22756.286 So, According to my answer ∆p amount of vapour pressure will be low. But my answer and book answer is not same. I don't want solution of this question. You just tell me the place where I am going in wrong direction in my solution. |
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Answer» Lowering of vapour pressure due to solute in 1 molal aqueous solution at 100°C is BOOK ANSWER - 13.44 Torr MY SOLUTION Since molality is 1 mole kg-1 That's mean 1 mole of solute is present in 1 kg of solvent (H2O) No. Of moles of Water = 1000g/18 = 55.55 No. Of moles of solute = 1 Mole Fraction of solute = 1/1+55.55 = 0.0176 .... (I) Pressure of solvent (assuming ideal behaviour) PV = nRT P = 55.55 x 0.0821 x 373 / 1 (d of water = 1) P = 1701.2774 atm 1 atm = 760 Torr psolvent = 1292970.8 Torr We know from Raoult Law ∆p = Xsolute . P(solvent) ∆p = 0.0176 x 1292970.8 ∆p = 22756.286 So, According to my answer ∆p amount of vapour pressure will be low. But my answer and book answer is not same. I don't want solution of this question. You just tell me the place where I am going in wrong direction in my solution. |
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