Chemistry

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I am having trouble with this question: A byproduct is chloroform (CHCL3) , a suspected carcinogen produced when HOCL, formed by the reaction of CL2 and water, reacts with dissolved organic matter. If a solution consist of 100 ppb of CHCL3 in drinking water, convert this concentration into molarity, molality, mole fraction,and mass percent. The main problem I have is getting the mass of CHCL3 and of the solvent in which in this case is water. Thank You!!!

  • Chemistry -

    I have no idea how to do this problem, but about the mass of CHCl3, don't you just add the atomic mass up from each element to find its total mass?

  • Chemistry -

    you don't need mass directly, you are given 100ppb

    molality= 100E-6 grams/molmassCHCl3*1kg

    remember you need the top mass in grams, not kg (1000E-9ppb*1kg)

    noticed the denominator ignored the solute, it is so dilute. Now in molarity, ignore the solute in the volume, and assume denstity is 1.00

    molarity=100E-6/molmass*1liter

    and so on.

    Chopsticks, you are right, you have no idea on this.

  • Chemistry -

    100 ppb means 100 parts per billion. That's the same type problem as percent by mass but we are dealing with billions instead of hundreds (percent is hundreds).
    So 100 ppb is 100 grams/1,000,000,000 g solution or 100 grams/10^9 grams solution which also means 100 grams/(100 g CHCl3 + 999,999,900 g solvent). You can see that the 100 is so small in comparison to the billion, that we can just ignore that difference for this part of the problem and say we have 100 grams/10^9 g solvent. So 100 g is how many moles? 100/molar mass CHCl3.
    Then moles/liter (and the density will be so close to 1.00 g/mL that you can convert grams solvent to mL solvent. After you have molarity, I think the rest will follow for you.
    100/119.4 = 0.8375 moles CHCl3/10^9 mL or
    0.8375 x 10^-9 moles/mL or
    0.8375 x 10^-9 moles/mL x(1000 mL/L) = 0.8375 x 10^-6 moles/L =
    8.375 x 10^-7 moles/L which rounds to 8.38 x 10^-7 Molar CHCl3 to 2 significant figures (the 100 has two s.f.). I would call that a very dilute solution. Check my work.

  • Chemistry -

    13.4 Expressing Solute Concentration
    • Concentration – the ratio of the quantity of
    solute to the quantity of solution (or solvent)
    ��Molarity (M) – the number of moles of solute
    per 1 liter of solution
    M = (mol of solute)/(liters of solution)
    – M is affected by temperature (the solution volume
    changes with T, so M changes too)
    – The solution volume is not a sum of the solvent and
    solute volumes (it must be measured after mixing)
    ��Molality (m) – the number of moles of solute per
    1 kilogram of solvent
    m = (mol of solute)/(kilograms of solvent)
    – m is not affected by temperature (the amounts of
    solute and solvent don’t change with T)
    – The solution volume is not needed; m can be
    calculated from the masses of solute and solvent
    • M and m are nearly the same for dilute aqueous
    solutions since 1 L of water is about 1 kg, so (liters
    of solution) ≈ (kg of solvent)
    Example: Calculate M and m for a solution
    prepared by dissolving 2.2 g of NaOH in 55 g of
    water if the density of the solution is 1.1 g/mL.
    m (molal )
    mol solute
    1.0 m
    kg
    1.0mol
    0.055 kg water
    0.055 mol NaOH
    0.055 mol
    40 g NaOH
    2.2 g NaOH 1 mol NaOH
    = = →
    = × =
    M (molar )
    density
    Volume mass
    1.1 M
    L
    1.1mol
    0.052 L solution
    0.055 mol NaOH
    52 mL
    1.1 g/mL
    2.2 g 55 g
    = = →
    =
    +
    = =
    • Parts of solute by parts of solution
    ��Parts by mass
    – Mass % – grams of solute per 100 grams of
    solution → % (w/w)
    100%
    mass of solution
    Mass%= mass of solute ×
    100%
    volume of solution
    Volume%= volume of solute ×
    – ppm or ppb – grams of solute per 1 million or 1
    billion grams of solution (for trace components)
    ��Parts by volume
    – Volume % – volume of solute per 100 volumes
    of solution → % (v/v)– ppmv or ppbv – volume of solute per 1 million
    or 1 billion volumes of solution (used for trace
    gases in air)
    ��Mole fraction (X) – ratio of the # mol of
    solute to the total # mol (solute + solvent)
    mol of solute mol of solvent
    mol of solute
    +
    X =
    Example: Calculate the X of NaOH in a solution
    containing 2.2 g of NaOH in 55 g of water.
    0.018
    0.055 mol 3.1 mol
    0.055 mol
    3.1 mol H O
    18 g
    0.055 mol NaOH 55 g 1 mol
    40 g
    2.2 g 1 mol 2
    =
    +
    =
    = =
    X
    • Converting units of concentration
    Example: A sample of water is 1.1×10-6 M in
    chloroform (CH3Cl). Express the concentration
    of chloroform in ppb. (Assume density of 1.0
    g/mL)
    1.1×10-6 M → 1.1×10-6 mol CH3Cl per 1 L solution
    10 ppb 56 ppb
    1.0 10 g
    5.6 10 g CH Cl
    1.0 10 g
    mL
    1 L 1000 mL 1.0 g
    5.6 10 g CH Cl
    1 mol CH Cl
    1.1 10 mol CH Cl 50.5 g CH Cl
    9
    3
    3
    5
    3
    3
    5
    3
    3
    3
    -6
    × =
    ×
    ×
    → × = ×
    × × = ×


    solution
    solution
    Example: What is the molality of a solution of
    methanol in water, if the mole fraction of
    methanol in it is 0.250?
    Assume 1 mol of solution:
    → nmeth = 1 mol × 0.250 = 0.250 mol
    → nwater = 1 - 0.250 = 0.750 mol
    18.5 m
    0.0135 kg H O
    0.250 mol methanol
    0.0135 kg H O
    10 g
    1 kg
    1 mol H O
    0.750 mol H O 18.0 g H O
    2
    3 2
    2
    2
    2
    = =
    × × =
    m
    Example: What is the molality of a 1.83 M
    NaCl solution with density of 1.070 g/mL?
    Assume 1 L (103 mL) of solution:
    → nNaCl = 1.83 mol
    1.90 m
    0.963 kg H O
    1.83 mol NaCl
    1070 g 107 g 963 g 0.963 kg
    107 g
    1 mol
    a 1.83 mol 58.44 g NaCl
    1070 g
    1 mL
    10 mL 1.070 g
    2
    3
    = =
    = − = =
    = × =
    = × =
    m
    mass of water
    mass of N Cl
    mass of solutionExample: What is the molarity of a 1.20 m KOH
    solution in water having density of 1.05 g/mL?
    Assume 1 kg (1000 g) of solvent (H2O):
    → nKOH = 1.20 mol
    1.18 M
    1.02 L
    1.20 mol KOH
    1016 mL 1.02 L
    1.05 g
    1067 g 1 mL
    1000 g 67.3 g 1067 g
    67.3 g
    1 mol
    1.20 mol 56.1 g KOH
    = =
    = × = =
    = + =
    = × =
    solution
    M
    volume of solution
    mass of solution
    mass of KOH

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