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What is the molality of 1M NH4OH in 70.092ml of water?

  1. Aug 11, 2017 #1
    1. The problem statement, all variables and given/known data
    What is the molality of 1M NH4OH dissolved in 70.092ml of water? What is the total mass of the solution? What is the total volume? What is the concentration percent? What is the molar concentration?

    1M NH4OH = 35.046g
    density of water = 1g/ml
    density of NH4OH = 0.88 g/ml

    2. Relevant equations
    molality = moles of solute/kg of solvent

    mass = volume x density

    molarity = moles of solute / liters of solution

    concentration = moles / volume

    3. The attempt at a solution
    Solve for molality:
    convert volume (ml) to mass (kg)
    70.092ml x 1g/ml =70.092g / 1000 = 0.07092kg
    1 / 0.070092 = 14.3 molals/kg

    Solve for total mass:
    35.046g + 70.092g = 105.138g

    Solve for total volume:
    convert mass(g) to volume(ml)
    35.046g / 0.88g/ml = 39.825ml
    39.825ml + 70.092ml = 109.917ml

    Solve for concentration percent:
    35.046g / 109.917ml=0.3188 x 100 = 31.88406= 32%


    Solve for molar concentration:
    1M / 0.109917L = 9.09777M/L = 9.1M/L
     
  2. jcsd
  3. Aug 11, 2017 #2

    Borek

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    Staff: Mentor

    I am afraid this question doesn't make much sense. First, you can't dissolve "1 M" in anything, "1 M" is a concentration, result of already dissolving something. Chances are what you mean is "What is the molality of solution prepared by dissolving 1 mole of NH4OH in 70.092ml of water". But even then it is still wrong. Dissolving 1 mole of ammonia (NH3), would be OK. Diluting ammonia solution (which is sometimes thought of as it was solution of NH4OH), would be OK. But there is no way of dissolving NH4OH, there is no such compound that you can isolate.
     
  4. Aug 11, 2017 #3
    My mistake. I should have treated this as an ammonia solution. I'll rewrite it asking for the molality of 1 mole NH3 dissolved in 70.092ml of water.
     
  5. Aug 12, 2017 #4

    Borek

    User Avatar

    Staff: Mentor

    Beware: the 0.88 g/ml density is also used incorrectly. It is density of the commercial solutions of ammonia (which are about 35%). To calculate concentrations precisely you need to find the density of your solution in tables.
     
  6. Aug 12, 2017 #5
  7. Aug 12, 2017 #6

    Borek

    User Avatar

    Staff: Mentor

    You need to find the density of the solution you are working with. Not the density of the gas, not the density of the stock solution, these are not things you are working with.

    Calculate the concentration (you have enough information to calculate at least molality and w/w %) and check the density of your solution in the tables.
     
  8. Aug 12, 2017 #7
  9. Aug 13, 2017 #8

    Borek

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    Staff: Mentor

    Nothing wrong with not knowing and trying to learn so no need to apologize. IMHO wikihow gives a poor (in a pedagogical sense) example of using 3 mL of a powdered solid - that's not how you do it in reality (you measure volume of solutions, or you weigh solids), plus, it leaves people confused (just like you are).

    Finding the density in tables is rather trivial. As I said - first, calculate the % w/w concentration of the SOLUTION (not a problem as you know mass of the solute and mass of the solvent). Then just look in the density tables, like this one:

    http://www.chembuddy.com/?left=CASC&right=density_tables

    or one that is more precise:

    https://wissen.science-and-fun.de/chemistry/chemistry/density-tables/density-of-ammonia/

    (if there is no exact concentration that you have calculated just take the closest one, or estimate the needed value with a linear approximation). For diluted solutions you can assume their density to be just that of pure water, 1 g/mL.

    Once you know the SOLUTION density (edit: and in this particular case the mass you have already calculated) you can calculate the SOLUTION volume, thus it will allow you to calculate volume based concentrations, like molarity or g/L (molality and w/w % don't require knowledge of the solution volume, as they are solely mass based).
     
    Last edited: Aug 13, 2017
  10. Aug 13, 2017 #9
     
  11. Aug 13, 2017 #10
    Thank you so much for your help! I've reworked the problem now.

    1. The problem statement, all variables and given/known data

    What is the molality of 1 mole of NH3 gas dissolved in 70.092ml of water? What is the total mass of the solution? What is the total volume? What is the concentration percent? What is the molar concentration?


    1mol NH3 = 17.031g


    2. Relevant equations

    molality = moles of solute / kg of solvent


    mass of solution = grams of solute + grams of solvent


    molarity = moles of solute / liters of solution


    volume = mass / density


    concentration = moles / volume


    mass percent of solution (w/w%) = mass of solute/mass of solution x 100


    3. The attempt at a solution



    Solve for molality:

    convert volume (ml) to mass (kg)

    70.092ml x 1g/ml =70.092g / 1000 = 0.07092kg

    1 / 0.070092 = 14.26 molals/kg



    Solve for total mass:

    17.031g + 70.092g = 87.123g



    Solve for total volume:

    Find w/w%

    17.031g / 87.123g = 0.19548 x 100 = 19.548 = 20 w/w%


    Look up density table for 20 w/w% NH3 solution

    density = 0.8951


    87.123g / 0.8951g/ml = 97.333ml


    Solve for concentration %:

    17.031g / 97.333ml = 0.1749 x 100 = 17.49 = 17.5 wt/vol %


    Solve for molar concentration:

    1 mol / 0.097333L= 10.274M (mol/L)
     
  12. Aug 14, 2017 #11

    Borek

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    Staff: Mentor

    Looks good :smile:

    (Actually turned out there was a mistake in the density table at the chembuddy site, so you are a bit off in your numbers, but the reasoning is perfectly OK).
     
    Last edited: Aug 14, 2017
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