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
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Definition 1 / 23
amount of solute in moles per liter of solution, M = moles (solute) / L (solution)
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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
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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
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