ap chemistry

a solution is prepared by mixing 50.0 mL toluene (C6H5CH3 d=0.867 g/mL) with 125 mL Benzene (C6H6 d=0.874 g/mL). assuming that the volumes add upon mixing, calculate the mass percent, mole fraction, molality, and molarity of the toluene.

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  1. http://www.jiskha.com/display.cgi?id=1420516971#1420516971.1420518334

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  2. but how u do it explain please

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  3. i don't know how to do it

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  4. I will get you started with mass percent.
    mass percent = (grams solute/mass solution)*100 = %

    What mass benzene do you have? Use density to calculate that.

    What mass toluene do you have? Use density to calculate that.

    mass solution - mass toluene + mass benzene.

    Now Substitute into mass percent equation above so that
    % toluene = (mass toluene/mass solution()*100 = ?

    For the others, lean on the definitions which are as follows:
    M = molarity = mols/L solution
    m = molality = mols/kg solvent
    Xtoluene = mols toluene/total mols
    mass = volume x density

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  5. mass of benzene is 0.874 g/ml

    mass of toluene= 0.867 g/mL

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  6. is this right: mass percent: toluene 28.4 %

    benezene: 7.6%

    mole fraction x C6H5CH3: .251

    mole fraction of X C6H6 1.619

    molality: 2.3 molal
    molarity: 1.43 M is that right? idk

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  7. need help

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  8. I have bold faced the typo sentence below.

    I will get you started with mass percent.
    mass percent = (grams solute/mass solution)*100 = %

    What mass benzene do you have? Use density to calculate that.

    What mass toluene do you have? Use density to calculate that.

    mass solution = mass toluene + mass benzene.

    Now Substitute into mass percent equation above so that
    % toluene = (mass toluene/mass solution()*100 = % toluene

    For the others, lean on the definitions which are as follows:
    M = molarity = mols/L solution
    m = molality = mols/kg solvent
    Xtoluene = mols toluene/total mols
    mass = volume x density

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  9. i figureed is this right DrBOb22

    is my ansswers right

    mass percent: 43.359/152.6gx100= 28.4%

    mass of benzene: 25 mLx .874 g/mL= 109.25 grams

    mass of touluene: 50 mLx .867 g/mL= 43.35 grams

    molarity: .251 mol/.175 L= 1.43 M

    molality: .251 mol/ .10925 kg= 2.30 m
    x touluene= .470 mol/1.87 mol= .251 mol

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  10. HELP ME WITH THIS ONE WHEN DONE WITH THE OTHER THX
    calculate the concentration of CO2 in a soft drink after the bottle is opened and equilibrates at 25C under a CO2 partial pressure of 0.0003 atm.

    c=KP henry's law

    k=?

    P=0.0003 atm. how to solve

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  11. is that right answer

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  12. 28.4% tolune is right.

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  13. what else is right help me

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  14. its touluene

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  15. what else is right?

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  16. The problem doesn't ask for % benzene.
    The mole fraction toluene is right (but I obtained 0.252 when rounded).
    mole fraction benzene is not but problem doesn't ask for that. However, it is 1-mols fraction toluene = 1-0.252 = ?

    M toluene = mols toluene/L soln and soln is 50 mL + 125 mL = 175 mL or 0.175L
    Then mol toluee = 43.35/92 = 0.471 so
    M = 0.471/0.175 = ?
    They don't ask for M benzene but it's done the same way. mols benzene/0.175 = ?
    m = mols/kg solvent = 0.471/kg solvent.
    mass is approx 153 so m = about 0.471/0.153 = about 3.08

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  17. k for CO2 is 3.4E-2 mols/L*atm
    c=KP
    Substitute p of 3.4E-2 for K and 3E-4 for P, solve for C (in M or mols/L)

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  18. a solution is made by dissolving 25 g NaCl in enough water to make 1.0 L solution. Assume that density of the solution is 1.0 g/mL. calculate the mass percent, molarity, molality, and mole fraction of NaCl.

    help please

    25g/58.44g NaCl=.428 mol

    1000 grams of solvent H20

    25g/58,44 g x100= 42.8 % mass percent

    molarity .428 mol/1.0 L= .428 M

    molality= .428 mol/1 Kg= .428 molal is that right?

    help please

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  19. is that right please help me

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  20. 25g/58.44g NaCl=.428 mol

    1000 grams of solvent H20

    25g/58,44 g x100= 42.8 % mass percent
    No, you have 25 g NaCl in 1000g which is 2.5 g/100 so it is 2.5%.

    molarity .428 mol/1.0 L= .428 M

    molality= .428 mol/1 Kg= .428 molal is that right?
    The M and m are right.

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  21. thanks good night

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  22. 25g/58.44g NaCl=.428 mol

    1000 grams of solvent H20

    25g/58,44 g x100= 42.8 % mass percent

    molarity .428 mol/1.0 L= .428 M

    molality= .428 mol/1 Kg= .428 molal is that right?

    help please
    ap chemistry - an, Tuesday, January 6, 2015 at 12:47am
    is that right please help me
    ap chemistry - DrBob222, Tuesday, January 6, 2015 at 12:47am
    25g/58.44g NaCl=.428 mol

    1000 grams of solvent H20

    25g/58,44 g x100= 42.8 % mass percent
    No, you have 25 g NaCl in 1000g which is 2.5 g/100 so it is 2.5%.

    molarity .428 mol/1.0 L= .428 M

    molality= .428 mol/1 Kg= .428 molal is that right?
    The M and m are right.

    the answer my teacher gave for the molality was .439 m not .428 m

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  23. also for the above question the answer for the mole fraction of NaCl is 0.00785 not 0.00766 what i do wrong

    moles of NaCl (.428 mol)/ moles of solution (55.9) = 0.00766 what i do wrong?

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  24. HELP ME WITH THIS ONE WHEN DONE WITH THE OTHER THX
    calculate the concentration of CO2 in a soft drink after the bottle is opened and equilibrates at 25C under a CO2 partial pressure of 0.0003 atm.

    c=KP henry's law

    k=?

    k for CO2 is 3.4E-2 mols/L*atm
    c=KP
    Substitute p of 3.4E-2 for K and 3E-4 for P, solve for C (in M or mols/L)
    P=0.0003 atm. how to solve

    not getting answer of 9.3E-6 M

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  25. Yes, and I missed that.
    The solution has a mass of 1000 g if the density is 1.0 g/mL but 25g of that is NaCl so the mass solvent is not 1 kg (as I led you to believe) but 1000g-25g = 975g or 0.975 kg.
    Then m = mols/kg solvent = 0.428/0.975 = ?
    I have answered above that both M and m are right and I'm not going to change all of those duplicate questions and responses. Hope this taks care of it.

    I hope this shows you how to get responses. If you show your work we can find the errors faster.

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  26. For the (CO2) I used 3.4E-2 for K in mol/L*atm and 29.41 for K in L*atm/mol (just the reciprocal). Your teacher may have used a different K value. Mine came from Wikipedia. A K of 3.25 mol/L*atm will give you 9.75E-6M

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  27. tanks

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