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Arihangdu

  • one year ago

When a 2.5 mol of sugar(C11H22O11) 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? Please show me all calculation.

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  1. abb0t
    • one year ago
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    z0mg fan + medal?! U promis?!

  2. Arihangdu
    • one year ago
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    Help me please

  3. Arihangdu
    • one year ago
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    Yes those who help will get

  4. cuanchi
    • one year ago
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    That is extortion!!!

  5. cuanchi
    • one year ago
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    delta T= k x molality the change in temperature is proportional at a k (constant) that depends of the nature of the solvent and the molality (concentration of the solute in the solution as moles of solute per kg of solvent. the sugar is producing one unit per mole when put it in water, but the aluminum nitrate 4 units per mole. Then the active molality of the aluminum nitrate will be 4 times higher than the sugar.

  6. Arihangdu
    • one year ago
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    Help please

  7. Arihangdu
    • one year ago
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    Help on doing maths please

  8. aaronq
    • one year ago
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    Have you tried setting up the problem? the formula is \(\sf \Delta T=molality*K_b\) where Kb is the boiling point elevation constant (for water) So first find the molality of the solution, the definition of molality is \(\sf molality=\dfrac{moles~of~solute}{Kg~of ~solvent }\)

  9. Arihangdu
    • one year ago
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    2.5 mole/2 kg solvents = 1.25 kg of solvent water. Al(NO3). ----> 3NO3 + Al^3+ We have 3 mol of Nitate and 1 mol of Al. Is this correct my friends?

  10. Arihangdu
    • one year ago
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    2.5 mole/2 kg solvents = 1.25 kg of solvent water. Al(NO3). ----> 3NO3 + Al^3+ We have 3 mol of Nitate and 1 mol of Al. Is this correct my friends?

  11. Arihangdu
    • one year ago
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    I stuck in this question its too hard for me please help me,,,!,

  12. cuanchi
    • one year ago
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    Yes that is correct. Then the Kb is the same for both conditions, the difference is the molality, you can said Delta T1/ molality 1 = Delta T2 / molality 2 Delta T2 = Delta T1 x molality 2 / molality 1 1 can be the sugar an 2 can be th aluminium nitrate The problem doesn't mention the mass of water (we don't know if it is one kilogram or any other mass) but doesn't matter because both, the sugar an the nitrate are dissolved in the same mass. The only important thing is that sugar doesn't dissociate in water so is producing one particle an the aluminium nitrate 4 particles. In place of molality we can use directly the amount of moles.. In the sugar solution we will have 2.5 mol an in the ammonium nitrate 2.5 x 4 = 10 Delta T2= 1 x 10 / 2.5 = 4 C !!!!! Easy as 1 x 4 = 4 that's all Folks

  13. Arihangdu
    • one year ago
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    Thank you very much.i appreciate your help.

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