1-3: A stock solution of 70% v/v ethanol was diluted with distilled water to prepare 1L of 21% v/v ethanol. (Density of ethanol =0.789 g/mL; MW of ethanol = 46 g/mol)1.Determine the initial volume of 70% v/v ethanol used in the preparation.2.How much water was added to the stock solution to make 1L of the final concentration?3.The molarity of 5.0 mL of 70% v/v ethanol was found to be4-6: A solution contains 45.0 g of methanol, CH3OH, dissolved in sufficient water to give a total mass of 210.6 g.4. Compute for the number of moles of the solute present in the given solution.5. What is the mass (in kg) of water used to dissolve the solute from the given problem?6. Determine the molal concentration of the said solution.5-7: A 10.34 molal aqueous nitric acid solution was prepared.(Density of nitric acid = 1.51 g/mL, density of water = 1.0 g/mL, and the MW of nitric acid = 63 g/mol)5. How many grams of HNO3 were used to prepare the solution?6. What is the Molarity of the solution?7. What is the concentration (%"/,) of the solution?

Answered: Image /qna-images/answer/9f1870e9-1335-4f1f-9514-4b5a18f2314c.jpg

1-3: A stock solution of 70% v/v ethanol was diluted with distilled water to prepare 1L of 21% v/v ethanol. (Density of ethanol = 0.789 g/mL; MW of ethanol = 46 g/mol) 1. Determine the initial volume of 70% v/v ethanol used in the preparation. 2. How much water was added to the stock solution to make IL of the final concentration? 3. The molarity of 5.0 mL of 70% v/v ethanol was found to be

1-3: A stock solution of 70% v/v ethanol was diluted with distilled water to prepare 1L of 21% v/v ethanol. (Density of ethanol = 0.789 g/mL; MW of ethanol = 46 g/mol) 1. Determine the initial volume of 70% v/v ethanol used in the preparation. 2. How much water was added to the stock solution to make IL of the final concentration? 3. The molarity of 5.0 mL of 70% v/v ethanol was found to be

Chemistry: Principles and Reactions

8th Edition

ISBN:9781305079373

Author:William L. Masterton, Cecile N. Hurley

Publisher:William L. Masterton, Cecile N. Hurley

Chapter10: Solutions

Section: Chapter Questions

Problem 6QAP

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6-16 Answer true or false. (a) Solubility is a physical property like melting point and boiling point. (b) All solutions are transparent—that is, you can see through them. (c) Most solutions can be separated into their components by physical methods such as distillation and chromatography.
6-40 What is the molarity of each solution? (a) 47 g of KCI dissolved in enough water to give 375 mL of solution (b) 82.6 g of sucrose, C12H22O11, dissolved in enough water to give 725 mL of solution (c) 9.3 g of ammonium sulfate, (NH4)2SO4 dissolved in enough water to give 2.35 L of solution
6-15 Answer true or false. (a) A solute is the substance dissolved in a solvent to form a solution. (b) A solvent is the medium in which a solute is dissolved to form a solution. (c) Some solutions can be separated into their components by filtration. (d) Acid rain is a solution.
Patients undergoing an upper gastrointestinal tract laboratory test are typically given an X-ray contrast agent that aids with the radiologic imaging of the anatomy. One such contrast agent is sodium diatrizoate, a nonvolatile water-soluble compound. A 0.378-m solution is prepared by dissolving 38.4 g sodium diatrizoate (NaDTZ) in.l.60 102 mL water at 3 1.2C (the density of water at 31.2C is 0.995 g/cm3). What is the molar mass of sodium diatrizoate? What is the vapor pressure of this solution if the vapor pressure of pure water at 31.2C is 34.1 torr?
1-3: A stock solution of 70% v/v ethanol was diluted with distilled water to prepare 1L of 21% v/v ethanol. (Density of ethanol = 0.789 g/mL; MW of ethanol = 46 g/mol) Determine the initial volume of 70% v/v ethanol used in the preparation. How much water was added to the stock solution to make 1L of the final concentration? The molarity of 5.0 mL of 70% v/v ethanol was found to be __________.
Consider a 1.000-L stock solution of 9.69 M Na3PO4. A 2.000-mL aliquot is taken from this stock solution and diluted to a final volume of 1.000 L. A 5.000-mL aliquot is then taken from this new solution and further diluted to make a new solution with a final volume of 750.0 mL. Calculate Na+ in the final solution.
I had a 6.00 M NaOH solution and I diluted it to a 50mL solution of 0.300 M NaOh solution by adding 2.5 mL of NaOH and filled it up to the mark using DI water. I also had to make a 50mL of a 0.300 M NaCl solution by adding 0.8g of NaCl. I then pipet 20mL of the 0.300M NaOH solution into a small beaker and pipet 10.0mL of 0.300 M NaCl into it. What is the molarity/concentration of NaOH in the new solution I created by dilution? I need to use the formula M1V1=M2V2
1. A 50.00 mL stock CaCl2 solution was diluted with water in a 250.0-mL volumetric flask. A 25.00 mL aliquot of this was further diluted to a final volume of 100.0 mL. Finally, 20.00-mL aliquot of the resulting solution was analyzed and found to contain 0.1500 M CaCh2 Solve for the concentration of the stock solution in M. 2. A 50.00 mL stock CaCl2 solution was diluted with water in a 250.0-mL volumetric flask. A 25.00 mL aliquot of this was further diluted to a final volume of 100.0 mL. Finally, 20.00-mL aliquot of the resulting solution was analyzed and found to contain 0.1200 g CaCl2 Solve for %w/v CaCl2 in the stock solution.
I need to make a solution containing 0.025 grams per liter. However, I do not want to weigh 0.025 grams and put this in one liter of solvent. I want to make a stock solution of 0.25 grams in 500 ml of solvent and take a unknown aliquot and dilute this in 500ml of solvent instead of 1000 ml of solvent. How do I calculate the volume of aliquot from the stock solution to achieve a final concentation of 0.025 grams per liter? What is the proper equation(s) required to figure out the proper aliquots required?
If a solution is made by adding 8g of (NH4)2SO4 to a 100 ml volumteric flask and enough water is added to bring the total volume to 100ml, what would the density of the solution be? This is a straightforward question but I just wanted some clarification. When we bring the total volume to 100ml, does that mean we added 100ml of water? Will the mass of the water be 100 g regardless of the amount of solute added?
Use only 1 decimal point (X.X) for all atomic weights. What will be the final molarity of 10.0 mL of 12.0 M HNO2 dilluted to 500.0 mL?
1) Which of the two cooling curves, left or right, represents the cooling of a pure solvent and which represents the cooling of a solution with a solute? 2) Assuming the freezing point is where the two regression lines for each figure meet, at what time did each solution begin to freeze and what is the freezing point in Celsius for each sample? 3) What was the change in the freezing point due to the addition of the solute?