Preview

N3 Solution Stoichiometry And Molar Dilution Problems

Good Essays
Open Document
Open Document
531 Words
Grammar
Grammar
Plagiarism
Plagiarism
Writing
Writing
Score
Score
N3 Solution Stoichiometry And Molar Dilution Problems
1. What is the molarity of a solution of ammonium chloride prepared by diluting 50.00 mL of a 3.79 M NH4Cl solution to 2.00 L?
2. A student takes a sample of KOH stock solution and dilutes it with 100.0 mLof water. The student determines that the diluted solution is 0.046 M KOH, but has forgotten to record the volume of the original stock solution sample. The concentration of the stock solution is 2.09 M. What was the volume of the original sample of stock solution?
3. A chemist wants to prepare a stock solution of H2SO4 (sulfuric acid), so that a sample of 20.00 mL will produce a solution with a concentration of 0.50 M when added to 100.0 mL of water.
A) What should the molarity of the stock solution be?
B) If the chemist wants to prepare 5.00 L of the stock solution from concentrated H2SO4, which is 18.0 M, what volume of concentrated acid should be used?
4. To what volume should 1.19 mL of an 8.00 M acetic acid solution be diluted in order to obtain a final solution that is 1.50 M
5. What volume of a 5.75 M solution be used to prepare 2.00 L of a 1.00 M solution?
6. A 25.00 mL of ammonium nitrate solution produces a 0.186 M solution when diluted with 50.00 mL of water. What is the molarity of the stock solution?
7. What mass of NaCl is required to precipitate all the Ag1+ ions from 20.0 mL of 0.100 M AgNO3 solution?
8. What mass of NaOH is required to precipitate all the Fe2+ ions from 50.0 mL of 0.200 M Fe(NO3)2 solution?
9. If 25.8 mL of AgNO3 solution is required to precipitate all of the Cl ions in a 0.785-g sample of KCl (forming AgCl), what is the molarity of the AgNO3 solution?
10. If 55.8 mL of BaCl2 solution is needed to precipitate all the sulfate in a 0.544-g sample of Na2SO4 (forming BaSO4), what is the molarity of the BaCl2 solution?
11. In the laboratory, 6.67 g of Sr(NO3)2 is dissolved in enough water to form a volume of 0.750 L. A 0.100-L sample of this stock solution is mixed with a 0.0460 M solution of Na2CrO4. What volume of Na2CrO4

You May Also Find These Documents Helpful

  • Good Essays

    What volume of 0.10 M NaOH can be prepared from 250. mL of 0.30 M NaOH?…

    • 581 Words
    • 3 Pages
    Good Essays
  • Satisfactory Essays

    Molar Mass Lab

    • 478 Words
    • 2 Pages

    also have 1.64 g / 18.22 g of solvent = 1.64 x 1000 / 18.22 g / kg = 90.0 g…

    • 478 Words
    • 2 Pages
    Satisfactory Essays
  • Satisfactory Essays

    LAB 20C

    • 561 Words
    • 3 Pages

    3. Molarity of HCl = 5.26 x 10-3 mol / 0.010 L = 0.526 M…

    • 561 Words
    • 3 Pages
    Satisfactory Essays
  • Satisfactory Essays

    Lab41

    • 279 Words
    • 2 Pages

    In this lab we are trying to find the amount of Cl- in a sample we know that Ag+ will bond with Cl- with a 1 to 1 ratio to form a precipitate. When we go from grams of Ag+ to moles of Cl- we can find out how much Cl- was in the unknown…

    • 279 Words
    • 2 Pages
    Satisfactory Essays
  • Good Essays

    4. A solution is prepared by dissolving 2 g of KCl in 100 g of H2O. In this solution, H2O…

    • 1080 Words
    • 5 Pages
    Good Essays
  • Satisfactory Essays

    4. A 1 ml sample of bacteria was diluted to 10-6 and 100μl was plated on the Petri dish at right. What was the approximate concentration of the original sample?8 x 10^7 bacteria per mL.…

    • 250 Words
    • 1 Page
    Satisfactory Essays
  • Satisfactory Essays

    Chem Exam 2

    • 1533 Words
    • 7 Pages

    ____ 4. Calculate the molarity of the solution formed by diluting 50.0 mL of 0.436 M NH4NO3 to 250.0 mL.…

    • 1533 Words
    • 7 Pages
    Satisfactory Essays
  • Satisfactory Essays

    Given the mass of NaCl we can easily calculate moles using the fence method. First find the mass of one mole of NaCl, which is equal to 58.44 g. Then take the amount of moles you have been given, 1.7, and divide that by 58.44 g. The total number of moles in this problem is 0.03 moles. Given the mL of solution we can easily calculate liters, again using the fence method. Take the amount of mL given and use a conversion factor of 1 mL/1000 L to calculate that there is 0.75 liters in the final solution. To find Molarity you divide 0.03 moles/ 0.75 liters to end up with 0.04 Molarity.…

    • 1243 Words
    • 5 Pages
    Satisfactory Essays
  • Satisfactory Essays

    18. You need to prepare 25.0mL of a solution of 1.5M HCl. Your lab provides you with 6M HCl stock solution. How do you prepare the solution you need?…

    • 464 Words
    • 2 Pages
    Satisfactory Essays
  • Good Essays

    3. Prepare a 20 mEq dose of a medication. Available solution is 40 mEq/10 ml. How many ml will be administered?…

    • 609 Words
    • 3 Pages
    Good Essays
  • Satisfactory Essays

    4. A drug is available in powdered form. It must be reconstituted by adding 10 ml sterile water to the vial. The resulting concentration is 125 mg per mL. The patient is to receive 200 mg. Give the patient __1.6_____mL.…

    • 310 Words
    • 2 Pages
    Satisfactory Essays
  • Satisfactory Essays

    Doc2

    • 253 Words
    • 2 Pages

    6. For a chemistry experiment Laura needs 300 ml of a 15% sulphuric acid solution. The Science Department has quantities of 5% acid solution and 20% acid solution. What volume of each solution is required to mix together to get her required concentration?…

    • 253 Words
    • 2 Pages
    Satisfactory Essays
  • Satisfactory Essays

    2. 100.00 g of sodium chloride is dissolved in enough water to prepare 10.0 L of solution.…

    • 416 Words
    • 2 Pages
    Satisfactory Essays
  • Satisfactory Essays

    Assuming complete reaction, what volume of 0.200 mol dm–3 potassium hydroxide solution (KOH(aq)), is required to neutralize 25.0 cm3 of 0.200 mol dm–3 aqueous sulfuric acid, (H2SO4(aq))?…

    • 678 Words
    • 3 Pages
    Satisfactory Essays
  • Satisfactory Essays

    How to Maker Solutions

    • 1042 Words
    • 5 Pages

    =190g/l 250ml 0.05M = 4.762g =0.1N | Succinic acid HO2CCH2CH2CO2H, FW = 118.09, Eq. = 59.045g/l 250ml 0.05M = 1.475g = 0.1N | Sulphamic acid H2NSO3H, FW = 97.09, Eq. = 99.09g/l 250ml 0.10M = 2.425 = 0.1N | Sulphuric acid H2SO4, FW = 98.08, Density = 1.8 1M = 56mls =…

    • 1042 Words
    • 5 Pages
    Satisfactory Essays