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anonymous

  • one year ago

When a 2.5 mol of sugar (C12H22O11) are added to a certain amount of water the boiling point is raised by 1 Celsius degree. If 2.5 mol of aluminum nitrate is added to the same amount of water, by how much will the boiling point be changed? Show all calculations leading to your answer OR use 3 – 4 sentences to explain your answer.

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  1. Photon336
    • one year ago
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    \[DeltaT = iK _{b}m \] i = the vast hoff factor which is how many moles of ions are dissolved. m = molality moles of solute/kilogram of solution i believe. Kb = a constant what your solvent is.

  2. Photon336
    • one year ago
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    I'm not completely sure but i'll do my best. 1. know the moles of sucrose 2. we know the temperature change 3. we know Kb we must find molality So that would be deltaT/(Kb*i) = m \[(1)/(0.52)(1) = 2 = m \] we know that we used 2.5 moles and molality is moles of solute/kg solvent 2 = 2.5mol/kg solute we do some math and we got 1.25 kg of solvent H2O \[Al(NO3_{3}) ---> 3NO _{3} + AL ^{3+}\] i'm not sure but you need to figure out how many moles of ions that are dissolved. so you have 3 moles of the nitrate and 1 mole of the aluminum so 3+1 = 4 for i I guess for the reaction with sugar that's 2.5 mol of sugar. and i = 1 but you're adding this to the other solution so i think you would need to take into account the sucrose as well. i = 5 kb =0.52 m = 5.0 moles/1.25 kg = 4 mol/kg ions. You would need to plug all these into your formula

  3. Photon336
    • one year ago
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    @Cuanchi I may have missed something

  4. cuanchi
    • one year ago
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    yea, dont get messed up, the problem is very simple the Kb is intrinsic of the solvent, and nothing to be with the solute. In this case the solvent in the two solution is the same (water) then the Kb is the same. The change of boiling point is proportional to the molality, and both solutions have the same number of moles in the same volume of water, both have the same molality. The van 't Hoff factor is different going to be different in the sugar you will have only one particle but in the Al(NO3)3 is going to produce 4 particles. The the change in the boiling point is going to be 4 times higher in the Al(NO3)3 than in the sugar solution.

  5. cuanchi
    • one year ago
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    @Photon336 I think the problem refers to two different solutions one with the sugar and the other with the Al(NO3)3 both in the same volume of water. But your calculations looks good if you think that they add the Al(NO3)3 to the original solution with the sucrose. https://answers.yahoo.com/question/index?qid=20150312113214AAt8p5P

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