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1000 g of 1m sucrose solution in water is cooled to-3. 5 degree centigrade, the weight of ice would be separated out at this temperature is

1000 g of 1m sucrose solution in water is cooled to-3. 5 degree centigrade, the weight of ice would be separated out at this temperature is

Grade:12th pass

1 Answers

Kunal ramchandani
13 Points
6 years ago
You`d extract 470 g of ice from this solution at that temperature.The approch here is to determine how much water remains in this solution after it`s cooled to that temperature and ice is formed.The equation for a solution`s freezing point depression isΔTf=Kf⋅b, whereΔTf - the freezing point depression, defined as the difference between the freezing point of pure water and the freezing point of the solution.Kf - the cryoscopic constant of water;b - the molalityYour solution`s freezing point depression would beΔTf=1.86∘Ckgmol−1⋅1molkg=1.86∘CSince your cooling temperature is almost twice as low as the solution`s normal freezing point, you can expect that close to 50% of the water is now ice.Since the molality of the solution was 1 molal at a total mass of 1000 g, you know that you have 1 mole of sucrose present.The mass of the water will decrease, which implies that the molality of the solution, which is expressed as moles of solute per liters of solution, will increase. This means that you can writeΔTf NEW=Kf⋅nsucrosex⋅10−3kg, wherex - the mass of the water expressed in grams. nsucrose - the number of moles of sucrose, in your case 1.Solve this equation for x to get3.534∘C=1.86∘C⋅mol−1⋅kg⋅1molex⋅10−3kgx=1.86⋅13.534⋅10−3=526.3 gThis is how much water your solution contains at that temperature. Thus, you can extractmice=msolution−mwatermice=1000−526.3=473.7 gI`ll leave the answer rounded to two sig figs, despite the fact that you only give one sig fig for the mass of the solution and for its molality.mice=470 g

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