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    Lec 5

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    The document provides information about calculating concentrations of solutions. It discusses the objectives of learning about solutions and concentration. It then defines key terms like molarity, molality, mole fraction, mass percent, volume percent, parts per million (ppm), and parts per billion (ppb). The document provides examples of calculating concentration using these various units and terms. It explains methods for determining concentration based on grams of solute and liters of solution, moles of solute and kilograms of solvent, and ratios of moles of solute to total moles.

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    Lec 5

    Uploaded by

    ibrahimomer922
    20 views35 pages
    The document provides information about calculating concentrations of solutions. It discusses the objectives of learning about solutions and concentration. It then defines key terms like molarity, molality, mole fraction, mass percent, volume percent, parts per million (ppm), and parts per billion (ppb). The document provides examples of calculating concentration using these various units and terms. It explains methods for determining concentration based on grams of solute and liters of solution, moles of solute and kilograms of solvent, and ratios of moles of solute to total moles.

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    LEC-5

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    The document provides information about calculating concentrations of solutions. It discusses the objectives of learning about solutions and concentration. It then defines key terms like molarity, molality, mole fraction, mass percent, volume percent, parts per million (ppm), and parts per billion (ppb). The document provides examples of calculating concentration using these various units and terms. It explains methods for determining concentration based on grams of solute and liters of solution, moles of solute and kilograms of solvent, and ratios of moles of solute to total moles.

    Copyright:

    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd
    Download as pdf or txt
    20 views35 pages

    Lec 5

    Uploaded by

    ibrahimomer922
    The document provides information about calculating concentrations of solutions. It discusses the objectives of learning about solutions and concentration. It then defines key terms like molarity, molality, mole fraction, mass percent, volume percent, parts per million (ppm), and parts per billion (ppb). The document provides examples of calculating concentration using these various units and terms. It explains methods for determining concentration based on grams of solute and liters of solution, moles of solute and kilograms of solvent, and ratios of moles of solute to total moles.

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    of 35

    Lecture Objectives

    By the end of this lesson the student is expected……….


    — Calculate concentration of the original bottle.
    — Calculate concentration of any solution .
    — Prepare a chemical compounds .
    — Express concentration in different methods .
    Solutions and Concentration
    — A solution consists of a liquid (the solvent) with a
    substance (the solute) dissolved in it. You are
    probably familiar with many solutions from your
    everyday life. Milk is a solution consisting of water
    (the solvent) with lactose and salts dissolved in
    it. Many of the solutions we work with in lab, are
    known as aqueous solutions because the solvent is
    water.
    Methods of Calculating Solution Concentration
    — There are number of ways to express the relative amounts
    of solute and solvent in a solution. we describe
    calculations for four different units used to express
    concentration:
    — Grams per liter
    — Percent Composition
    — Molarity
    — Molality
    — Normality
    — PPM
    — PPB
    Chemical concentrations

    Molarity = Moles of solute/Liters of Solution (M)

    Molality = Moles of solute/Kg of Solvent (m)

    Mole Fraction = Moles solute/total number of moles

    Mass % = Mass solute/total mass x 100

    Volume % = volume solute/total volume x 100

    ppm = parts per million *


    ppb = parts per billion *

    * mass for solutions, volume for gasses


    Chemical concentrations

    Molarity = Moles of solute/Liters of Solution (M)

    Molality = Moles of solute/Kg of Solvent (m)

    Mole Fraction = Moles solute/total number of moles

    Mass % = Mass solute/total mass x 100

    Volume % = volume solute/total volume x 100

    ppm = parts per million *


    ppb = parts per billion *

    * mass for solutions, volume for gasses


    Chemical concentrations

    Molarity = Moles of solute/Liters of Solution (M)

    Molality = Moles of solute/Kg of Solvent (m)

    Mole Fraction = Moles solute/total number of moles

    Mass % = Mass solute/total mass x 100

    Volume % = volume solute/total volume x 100

    ppm = parts per million *


    ppb = parts per billion *

    * mass for solutions, volume for gasses


    A sample of NaNO3 weighing 8.50 grams is placed in
    a 500. ml volumetric flask and distilled water was
    added to the mark on the neck of the flask. Calculate
    the Molarity of the resulting solution.
    1. Convert the given grams of solute to moles of solute :
    No of moles = Wt in grams = 8.50g NaNO3 = 0.1moleNaNO3
    M. wt 85 g NaNO3

    2. Convert given ml of solution to liters


    500 ml = 0.50 liter
    1000 ml

    3. Apply the definition for Molarity:


    Molarity = moles NaNO3 / volume of the solution in liters
    Chemical concentrations

    Molarity = Moles of solute/Liters of Solution (M)

    Molality = Moles of solute/Kg of Solvent (m)

    Mole Fraction = Moles solute/total number of moles

    Mass % = Mass solute/total mass x 100

    Volume % = volume solute/total volume x 100

    ppm = parts per million *


    ppb = parts per billion *

    * mass for solutions, volume for gasses


    Determine the molality of 3000. grams of solution containing 37.3 grams of
    Potassium Chloride KCl.

    1. Convert grams KCl to moles KCl


    37.3 g KCL = 0.50 mole KCL
    74.6 g KCL

    2. Determine the mass of pure solvent from the given grams of solution and
    solute
    Determine the molality of 3000. grams of solution containing
    37.3 grams of Potassium Chloride KCl.

    1. Convert grams KCl to moles KCl

    No of moles = Wt in grams = 37.3 g KCL = 0.50 mole KCL


    M. wt 74.6 g KCL

    2. Determine the mass of pure solvent from the given grams


    of solution and solute

    Total grams = 3000 grams = Mass of solute + Mass of solvent


     Mass of pure solvent = (3000 - 37.3) gram
    = 2962.7 gram
    .

    3. Convert grams of solvent to kilograms


    2962.7 grams solvent = 2.9627 kg
    1000 grams

    4. Apply the definition for molality


    Chemical concentrations

    Molarity = Moles of solute/Liters of Solution (M)

    Molality = Moles of solute/Kg of Solvent (m)

    Mole Fraction = Moles solute/total number of moles

    Mass % = Mass solute/total mass x 100

    Volume % = volume solute/total volume x 100

    ppm = parts per million *


    ppb = parts per billion *

    * mass for solutions, volume for gasses


    Determine the mole fraction of KCl in 3000. grams of solution
    containing 37.3 grams of Potassium Chloride KCl.

    1. Convert grams KCl to moles KCl

    37.3 g KCL = 0.50 mole KCL


    74.6 g KCL

    2. Determine the mass of pure solvent from the given grams of


    solution and solute

    Total grams = 3000 grams = Mass of solute + Mass of solvent

     Mass of pure solvent = (3000 - 37.3) gram


    = 2962.7 gram
    3. Convert grams of solvent H2O to moles

    2962.7 grams of water = 164.6 moles H2O


    18.0 grams

    4. Apply the definition for mole fraction mole fraction =


    Chemical concentrations

    Molarity = Moles of solute/Liters of Solution (M)

    Molality = Moles of solute/Kg of Solvent (m)

    Mole Fraction = Moles solute/total number of moles

    Mass % = Mass solute/total mass x 100

    Volume % = volume solute/total volume x 100

    ppm = parts per million *


    ppb = parts per billion *

    * mass for solutions, volume for gasses


    Determine the mass % of a NaCl solution if 58.5 grams of NaCl
    was dissolved in 50 ml of water (assume the density of
    water to be 1 g/ml)

    1. Convert ml of water to grams

    2. Determine total mass of solution

    Mass of solution = mass of solute + mass of solvent =


    58.5 + 50 = 108.5 g

    3. Apply the definition of mass percent mass % =


    58.5 / 108.5 *100 = 53.9% NaCl
    Chemical concentrations

    Molarity = Moles of solute/Liters of Solution (M)

    Molality = Moles of solute/Kg of Solvent (m)

    Mole Fraction = Moles solute/total number of moles

    Mass % = Mass solute/total mass x 100

    Volume % = volume solute/total volume x 100

    ppm = parts per million *


    ppb = parts per billion *

    * mass for solutions, volume for gasses


    Assuming the density of water to be 1 g/mL we
    approximate the density of a dilute aqueous
    solution to be 1 g/mL

     1 ppm = 1 µg/mL = 1 mg/L

     1 ppb = 1 ng/mL = 1 µg/L


    Determine the ppm of a NaCl solution if 58.5 grams of NaCl was dissolved in 50.0
    ml of water (assume the density of water to be 1 g/ml)

    Convert ml of water to grams

    Determine total mass of solution

    Mass of solution = mass of solute + mass of solvent =


    58.5 + 50.0 = 108.5 g

    Apply the definition of ppm

    58.5 / 108.5 * 106 = 5.39 x 105 ppm NaCl


    Features of an accurate quantitative process
    — The key features of an accurate quantitative
    process are as follows…
    — To perform an accurate quantitative analytical
    process first of all we need two thing, they are
    –
    — A representative sample &
    — An appropriate methodology.
    How Do We Express Concentrations of Solutions?
    — Molarity (M)= moles/liter or mmoles/mL
    — Normality(N) = equivalence/liter or meq/mL
    — Formality(F)= is identical to molarity
    — Molality(m) = moles/1000g solvent
    Expression of Analytical Results
    So Many Ways
    — Solid Samples:
    — %(wt/wt) = (wt analyte/wt sample)x 100%
    — ppm(wt/wt) = (wt analyte/wt sample)x 106 ppm
    — ppb(wt/wt) = (wt analyte/wt sample)x 109 ppb
    — Liquid Samples
    — %(wt/vol) = (wt analyte/vol sample mL)x 100 %
    — ppm(wt/vol) = (wt analyte/vol sample mL)x 106 ppm
    — ppb(wt/vol) = (wt analyte/vol sample,mL)x 109 ppb
    Example
    Percentage weight in sample
    — How many grams of dextrose are required to
    prepare 4000 ml of a 5% solution?
    — Solution
    — 5/100x4000=200g
    EXAMPLE
    Volume in Volume
    — How many milliliters of liquified phenol (2.5%)
    in 240 ml of Calamine lotion should be used in
    compounding prescription for eye drop?
    — Solution
    — Volume of liquified phenol=2.5/100x240=6ml
    Percent weight in weight
    — How many grams of phenol should be used to
    prepare 240 gram of a 5% (w/w) solution in water?
    — Solution
    — 5/100x240=12.0g
    Name Defining Units
    Molarity moles of solute/liter (solutions), or
    (e.g. 0.1200 M) millimoles/milliliter (solutions)

    Percent (e.g. 23.45 %) (grams of substance/grams of sample) x 100%

    milligrams/liter (solutions), or micrograms/milliliter


    Parts per million
    (solutions) milligrams/kilogram (solids), or
    (e.g 2.34 ppm, 2.34 mg/L)
    micrograms/gram (solids)

    Parts per billion micrograms/liter (solutions), or nanograms/gram


    (e.g. 0.45 ppb, 0.45 ug/L) (solids)
    Normality
    — Normality is equal to the gram equivalent weight of a
    solute per liter of solution. A gram equivalent weight or
    equivalent is a measure of the reactive capacity of a given
    molecule. Normality is the only concentration unit that is
    reaction dependent.
    — Equivalent Weight
    — One mole of a substance is defined as the weight (mass)
    in grams equal to its molecular weight (MW). Similarly,
    one equivalent of a substance is the weight in grams equal
    to its “equivalent weight” (EW).
    — Equivalent weight = MWt /n
    Normality
    — Normality(N) = equivalence/liter or meq/mL
    — For HCl, 1 mol HCl = 1 eq HCl. MW of HCl = 36.46;
    EW of HCl = 36.46.
    — For H2SO4, 1mol H2SO4 = 2 eq H2SO4. MW of
    H2SO4 = 98.08; EW of H2SO4 = 49.04.
    — For acids, EW = MW/n, where n is the number of
    acidic hydrogens in the chemical formula.
    — For hydroxide bases, EW = MW/n, where n is the
    number of hydroxide ions in the chemical formula
    — Calculate the number of equivalents in 220 grams of
    H3PO4
    — [the gram molecular weight of H3PO4 =98 gmw
    — [Divide the gmw by the valence (which is 3 due to the +3
    charge on the 3 H’s)
    — Gram per equivalence:
    — 98 g = 32.6 g/Eq
    — 3 eq
    — equivalents 220 g = 6.74 eq
    — 32.6 g/eq
    What is the normality of a solution containing 149
    grams of H2SO4 in 900 ml
    1) The gram molecular weight of H2SO4 =98 gmw
    2) Divide the gmw by the valence (which is 2 due to the
    +2 charge on the 2 H’s)
    3) Equivalent= 98 g = 49 g/Eq
    2 Eq
    Figure out how many equivalents are in 149 grams of
    H2SO4
    No. of equivalents= 149 g = 3.04 eq
    49 g/eq
    Normality = equivalents (Don’t forget to change ml to
    liters!!) liters
    Normality = 3.04 eq = 3.38 N
    0.9 L

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