College Chemistry

My professor gave us two homework questions that are a bit confusing, can you help me to get started on them.

1) You have 6g of water (Kf=1.86 C kg/mol). You add and dissolve 2.1 g of a different unknown substance. The freezing point of the solution is -0.3 deg C. What is the molecular mass of the solute?

2) A nitric acid solution has a density of 1.42g/ml and is 16 molar. What is its molality? What is the mole fraction of nitric acid? What is the weight percentage?

If you could help me out I would greatly appreciate it.

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  1. #1. This type problem illustrates the colligative properties of H2O. The freezing point of pure water is zero degrees C but a solute added makes the solution freeze lower than zero C.
    delta T = Kf*molality
    solve for molality.

    Then molality = moles/kg solvent.
    solve for moles.

    Then moles = g/molar mass
    solve for molar mass.

    #2. The HNO3 solution is 16 M. That means 16 mols HNO3/L of solution. The solution has a density of 1.42 g/mL; therefore, 1000 mL has a mass of
    1.42g/mL x 1000 mL = 1420 g.
    Part of that is water and part is HNO3. You know 16 moles of it is HNO3, the molar mass of HNO3 is approximately 63 (you should do it more accurately) and 16 moles would be 1008g; the amount of water must be 1420-1008 = ??
    Then molality = moles/kg solvent.
    Post your work if you get stuck.

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  2. I did both problems and these are my results:

    1) Molality concentration= -.161 kg/mol
    8.1g--> .0081 kg
    mols solute: (-.161kg/mol)(.0081kg)= -.0013 mol
    then, -.0013 mol H2O(18.016 g H2O)= -.0234 g H2O solute

    2) HNO3= 16 mols/l
    density= 1420 g
    16 mols HNO3(63.015)= 1008.2 g HNO3
    1420-1008= 412g
    Molality: 412g/1000g=.412 kg
    then, 16 mols/.412 kg= 38.8= molality

    mol fraction: (1.42g)(1000ml)=1420 g/l
    mols solvent: (1420g)/(63.015g/mol)= 22.5 mols solvent
    total # mols solution:
    =16 mol+22.5 mol=38.5 mols solution
    mol fraction: 16 mol/38.5= .4155

    weight percentage: mass solute/mass solution *100%
    38.5 mols*63.015g= 2426 g HNO3
    so, (1008g/2426g)*100%= 41.5% HNO3

    these are my solutions, can you check to see I didn't make any mistakes, thank you

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  3. Somewhere along the way you got mixed up on #1. The 0.161 m is correct but you go downhill from there.
    1) Molality concentration= -.161 kg/mol
    8.1g--> .0081 kg
    mols solute: (-.161kg/mol)(.0081kg)= -.0013 mol
    then, -.0013 mol H2O(18.016 g H2O)= -.0234 g H2O soluteI don't know why you are working on water---H2O is the solvent, not solute.
    0.161 m = moles/kg solvent.
    Kg solvent is 0.006 (6 g from the problem) so moles = 0.161/0.006 = 9.67 x 10^-4
    and moles = g/molar mass;
    molar mass = g/moles = 2.1/9.67 x 10^-4 = ??


    2) HNO3= 16 mols/l
    density= 1420 g
    16 mols HNO3(63.015)= 1008.2 g HNO3
    1420-1008= 412g
    Molality: 412g/1000g=.412 kg
    then, 16 mols/.412 kg= 38.8= molality
    OK to here.

    mol fraction: (1.42g)(1000ml)=1420 g/l
    mols solvent: (1420g)/(63.015g/mol)= 22.5 mols solvent
    The 1420 grams is the MASS of the SOLUTION (not the solvent). You worked the molality and calculated 0.412 kg solute above. You should have used 412 g solute here. I think if you correct that the rest will fall into line although you may need to recalculate each step
    total # mols solution:
    =16 mol+22.5 mol=38.5 mols solution
    mol fraction: 16 mol/38.5= .4155

    weight percentage: mass solute/mass solution *100%
    38.5 mols*63.015g= 2426 g HNO3
    so, (1008g/2426g)*100%= 41.5% HNO3

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