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    Performance Task #5: University of San Agustin

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    Micole Brodeth
    The document contains 7 chemistry problems involving calculations of percentage by mass and volume, mole fraction, molarity, molality, and parts per million (ppm) concentration. Each problem follows the format of stating given values, the calculation required, showing the step-by-step work, and providing the final answer. The problems cover a range of common solution stoichiometry calculations.

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    Performance Task #5: University of San Agustin

    Uploaded by

    Micole Brodeth
    129 views7 pages
    The document contains 7 chemistry problems involving calculations of percentage by mass and volume, mole fraction, molarity, molality, and parts per million (ppm) concentration. Each problem follows the format of stating given values, the calculation required, showing the step-by-step work, and providing the final answer. The problems cover a range of common solution stoichiometry calculations.

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    PT5Group2 (1)

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    The document contains 7 chemistry problems involving calculations of percentage by mass and volume, mole fraction, molarity, molality, and parts per million (ppm) concentration. Each problem follows the format of stating given values, the calculation required, showing the step-by-step work, and providing the final answer. The problems cover a range of common solution stoichiometry calculations.

    Copyright:

    © All Rights Reserved
    Download as DOCX, PDF, TXT or read online from Scribd
    Download as docx, pdf, or txt
    129 views7 pages

    Performance Task #5: University of San Agustin

    Uploaded by

    Micole Brodeth
    The document contains 7 chemistry problems involving calculations of percentage by mass and volume, mole fraction, molarity, molality, and parts per million (ppm) concentration. Each problem follows the format of stating given values, the calculation required, showing the step-by-step work, and providing the final answer. The problems cover a range of common solution stoichiometry calculations.

    Copyright:

    © All Rights Reserved
    Download as DOCX, PDF, TXT or read online from Scribd
    Download as docx, pdf, or txt
    of 7

    University of San Agustin

    General Luna St., 5000 Iloilo City, Philippines


    www.usa.edu.ph
    BASIC EDUCATION DEPARTMENT
    SENIOR HIGH SCHOOL

    PERFORMANCE TASK #5

    DIRECTION: Write your solution after each problem. Observe a systematic way of answering problems
    solving (Given, Required, Analysis, Solution). Final answer should be red and must follow the correct number
    of significant figure. 5 points each

    % by mass

    1. A solution is made by dissolving 125 g of sodium chloride in 1.5 kg of water. What is the percent by
    mass?

    Given:

    Solute (NaCl) = 125g

    Solvent (water) = 1.5 kg to g = 1500 g.

    Required:

    Calculate the percent by mass of NaCl solution.

    Solution:

    Percent mass =

    ( mass of solute/mass of solution) x 100%

    a. Find the mass of the solution.

    Solution= solute +solvent

    = 125g + 1500g

    = 1625 g

    %mass of NaCl = (125g/165g) x100%

    = 7.692%

    Answer:

    The percent mass of NaCl solution is 7.69%.


    % by volume

    1. What is the percent by volume of a solution formed by added 15 L of acetone to 28 L of water?

    Given:

    Solute (acetone) = 15L

    Solvent (water) = 28L

    Required:

    Calculate the percent volume of acetone solution.

    Solution:

    Percent volume =

    ( volume of solute/volume of solution) x 100%

    a. Find the volume of the solution.

    Solution= solute +solvent

    =15L + 28L

    = 43L

    %volume = (15L/43L) x100%

    = 34.88%

    Answer:

    The percent volume of the solution is 35%.


    Mole fraction

    1. A solution containing 75 grams of ethanol (C2H6O) in 450 grams in water.

    Given:

    Solute (ethanol, C2H6O) = 75g

    Solvent (water) = 450 g

    Required:

    Calculate the mole fraction.

    Solution:

    "mole fraction =

    mole of component /total mole of solution

    a. Calculate the moles of solute (ethanol, C2H6O)

    (75gx 1mol)/46 grams = 1.6304 moles

    b. Calculate the moles of water (450gx

    1mol)/18 grams = 25 moles

    c. Calculate the total moles of solution.

    1.6304 moles + 25 moles =26.6304 moles

    d. Calculate the mole fraction.

    mole fraction =

    mole of component /total mole of solution

    = 1.6304 moles/26.6304 moles

    = .0612

    Answer:

    The mole fraction of the solution is 0.0612.


    Molarity
    1.
    734 grams of lithium sulfate, Li2SO4 are dissolved to make 2500 mL of solution. What is the molar
    concentration?

    Given:

    The mass of lithium sulfate is 734 g.

    The volume is 2500 ml-2.5L.

    Required:

    Calculate the molar concentration.

    Solution:

    Molarity = number of moles (n)/volume (V)

    a. Calculate the molar mass of lithium sulfate (Li2SO4).

    =7(2)+32+16(4)

    = 110g/mol

    b.Calculate the number of moles.

    mole(n)= mass/molar mass

    =110g/mol/734 g

    = 6.67moles

    c. Calculate the molar concentration M

    = n/v

    6.67moles/2.5L =2.668mol/L

    Answer:

    The overall molarity of the solution is 2.67 mol/L


    Molality
    1.
    A 4.0-gram sugar cube C12H22O11 is dissolved in 350 mL teacup of 80˚ C water. What is the molality
    of the sugar solution?

    Given:

    Solute (C12 H22 O11) = 4 g

    Solvent (water) = 350 ml

    density 0.975g/mol in 80 °C water

    Required:

    Calculate the molality of the sugar solution.

    Solution:

    Molality = mole of solute/kg of solvent

    a. Calculate the moles of sucrose, C12 H22 O11

    =(12)(12) +(1)(22)+(16)(11)

    = 342g/mol

    = 4g/342g/mol

    = 0.0117mol

    b.Calculate the mass of the solvent (water) in kg.

    mass =( density )(volume)

    = 0.975g/molx 350ml

    = 341.25g

    =0.34125kg

    c. Calculate the molality


    Molality = mole of solute/kg of solvent

    =0.0117mol/0.34125kg

    =0.03428mol/kg

    Answer:

    The molality of the sugar solution is 0.034mol/kg.


    PPM

    1. Suppose 17 grams of sucrose dissolved in 183 grams of water. What is the concentration of sucrose in
    ppm?

    Given:

    Solute(sucrose)= 17g

    Solvent (water)= 183g

    Required:

    Calculate the concentration of sucrose in ppm.

    Solution:

    Cppm = (mass of solute/mass of solution) x106

    a. Determine the mass of the solution.

    solution = solute + solvent

    =17g+183g

    = 200g

    b. Calculate the ppm concentration.

    Cppm = (mass of solute/mass of solution) x 106

    = (17g/200g) X106

    =0.085X106

    Answer:

    The concentration of sucrose is 85,000 ppm.

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