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    calculate the number of moles of sulfuric acid that is contained in 0.250 ml of 0.500 m sulfuric acid solution.

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    Experiment 17: Concentration Units Flashcards

    Study with Quizlet and memorize flashcards containing terms like molarity, molality, % (m/m) and more.

    Experiment 17: Concentration Units

    Term 1 / 23 molarity

    Click the card to flip 👆

    Definition 1 / 23

    amount of solute in moles per liter of solution, M = moles (solute) / L (solution)

    Click the card to flip 👆

    Created by m_sanchez0512

    Terms in this set (23)

    molarity

    amount of solute in moles per liter of solution, M = moles (solute) / L (solution)

    molality

    amount of solute in moles per kilogram of solvent, m = moles (solute) / Kg (solvent)

    % (m/m)

    = 100% x grams / total mass

    WOTF concentration units are temperature dependent?

    molarity

    molarity is an example of

    volumetric unit volumetric unit

    value determined by volume

    molality and % (m/m) are examples of

    gravimetric units gravimetric unit value set by mass

    In titration labs, you used weak acid hydrogen phthalate (KHC8H4)4) as a primary standard in an acid/base titration "KHP", How much of this weak acid would have to be weighed out to prepare 100.0 mL of a 0.240 M KHP solution?

    ...

    If you actually weighed out 8.50% more KHP than required what would be the actual molarity of the solution?

    ...

    If you used the prepared KHP solution in a lab experiment which concentration would you need to use in your calculation? Explain your reasoning.

    ...

    You weight out 2.98 g of sodium sulfate and use it to create a 250-mL aqueous solution. What is the molarity of sodium ions in the resulting solution?

    ...

    Calculate the number of moles of sulfuric acid that is contained in 0.250 L of 0.500 M sulfuric acid solution

    Answer = 0.125 moles

    Explanation - 0.250 L x 0.500 M

    what mass of MgF2 is contained in 80.85 g of a 22.4% by mass solution of MgF2 in water?

    Answer = 18.1 g

    Explanation - 22.4% = 100% x (x grams) / 80.85 g

    sulfuric acid H2SO4 stock solution

    a solution that has been prepared with known molarity

    What is the molarity of a potassium bromide solution that is prepared by adding 400 mL of water to 25 mL of aqueous 2.00 M potassium bromide solution? Assume volumes are additive, report answer to three sig. figs.

    ...

    Part A - Preparation of 100.0 mL of an aqueous solution of sucrose, C12H22O11, with specified molarity

    Assigned Molarity = 0.0441 M

    Volume of Solution = 100.00 mL

    Mass of Sucrose = ? Answer = 1.51 g

    Explanation - 0.1 L x (0.0441 M / 1 L) x (342.308 g / 1 mol)

    Calculation - Mass of sucrose required

    Part B - Dilution of a 1.0 M NaCl solution to a specified final volume

    Assigned Final Volume = 76.2 mL

    Molarity of "Stock Solution" = 1.0 M NaCl

    Volume of "Stock Solution" = 10.0 mL

    Molarity of Final Solution = ?

    Answer = 0.13 M

    Explanation - (1.0 M NaCl x 0.01 L) / 0.07621 L = M2

    M1V1/V2 = M2

    Calculation - Find molarity of NaCl

    Part C - Preparation of 100.00 mL solution with a specified mass of alum, KAI(SO4)2.12H2O. Student will determine the molarity of a particular ion (K+, Al+3, SO4-2) in the resulting solution (K+, Al+2, SO4-2)

    Answer = 0.078 M Na Cl

    Explanation - 1.84 g Alum (1 mole/474.39 g Alum) = 3.8787x10^-3 mole

    3.8787x10^-3 x 2 = 7.7573 x 10^-3 mole

    7.7573 x 10^-3 / 0.1 L = 0.078 M NaCL

    Calculation - Molarity of ion in final solution

    Part D - Preparation of a sucrose solution at a specified % (m/m0 sucrose and a specified total mass of solution

    Assigned Mass of Solution = 35.3 g

    Assigned % by Mass Mass of sucrose = ? Mass of Water = ?

    Answers = 25.5 g Sucrose and 9.7 g H20

    Explanation - Suroces = 35.2 g x 0.725 (m/m)

    Water = 35.2 g - 25.5 g

    Calculation - Masses of sucrose and water required (dry lab)

    Part E - Preparation of a solution with specified molality of sucrose using a specified mass of water

    Assigned Molality = 0.339 g

    Assigned Mass of Water = 52.2 g

    Mass of Sucrose = ?

    Mass of Prepared Solution = ?

    Answer = Explanation -

    Calculation - Masses of sucrose required in solution (dry lab)

    Part F - Prepare a standard solution of KIO3 for Experiment 18

    Volume of Solution =

    Mass of KIO3 = Molarity of KIO3 = Answer = Explanation -

    Calculation - Molarity of prepared solution

    (Important to use deionized water, since solution will be used as a standard)

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    Calculate the number of moles of sulfuric acid that is contained in 0.250 mL of 0.500 M sulfuric acid

    Calculate the number of moles of sulfuric acid that is contained in 0.250 mL of 0.500 M sulfuric acid solution… Get the answers you need, now!

    01/30/2020 Chemistry High School

    answered • expert verified

    Calculate the number of moles of sulfuric acid that is contained in 0.250 mL of 0.500 M sulfuric acid solution.

    Is correct 0.25 ml? to add comment 00:00

    Expert-Verified Answer

    2 people found it helpful

    Ace 10.3K answers 37.7M people helped

    The number of moles of the sulfuric acid in the given new volume of the acid is 1.25 x 10⁻⁴ mole.

    The given parameters;

    volume of the sulfuric acid, V₁ = 0.25 ml

    concentration of the acid, C₁ = 0.5 M

    The concentration of the sulfuric acid is calculated as follows;

    Thus, the number of moles of the sulfuric acid in the given new volume of the acid is 1.25 x 10⁻⁴ mole.

    Learn more here:brainly.com/question/15659334

    5.0 (1 vote) to add comment Advertisement

    Answer

    1 person found it helpful

    Ace 7.2K answers 43.3M people helped Answer: 0.000125 moles Explanation:

    Firstly, we need to know the meaning of the 0.5M. This simply means 0.5 moles are present in 1L or 1dm^3. 0.5M could also be written as 0.5moles/L.

    Now, we know that there are 0.5 moles in 1L, we now need to calculate the number of moles in 0.5ml.

    It is necessary to know that 1000ml = 1L.

    Now if 0.5 moles = 1000ml

    xmoles = 0.25ml

    X = (0.25 * 0.5)/1000 = 0.000125 moles

    2.3 (3 votes)

    Need a bit more clarification? Get a high-quality answer with step-by-step explanations from a professional in just minutes instead!

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