work and use units throughout your calculations! 5. You wish to make 500. mL of 0.90 M KCl from solid KCl and DI water. a. How many liters of solution is this? 0.5 L b. How many moles of KCl would be in this volume of 0.90 M KCl? (hint: re-write the molarity in unabbreviated form first). 0.90 moles KCI IL c. What is the mass of that many moles of KCI?

work and use units throughout your calculations! 5. You wish to make 500. mL of 0.90 M KCl from solid KCl and DI water. a. How many liters of solution is this? 0.5 L b. How many moles of KCl would be in this volume of 0.90 M KCl? (hint: re-write the molarity in unabbreviated form first). 0.90 moles KCI IL c. What is the mass of that many moles of KCI?

Introductory Chemistry: A Foundation

8th Edition

ISBN:9781285199030

Author:Steven S. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Donald J. DeCoste

Chapter15: Solutions

Section: Chapter Questions

Problem 116AP: A large beaker contains 1.50 L of a 2.00 M iron(III) chloride solution. How many moles of iron ions...

See similar textbooks

Similar questions

Calcium carbonate, CaCO3, can be obtained in a very pure state. Standard solutions of calcium ion are usually prepared by dissolving calcium carbonate in acid. What mass of CaCO3 should be taken to prepare 500. mL of 0.0200 M calcium ion solution?
A large beaker contains 1.50 L of a 2.00 M iron(III) chloride solution. How many moles of iron ions are in the solution? How many moles of chloride ions are in the solution? You now add 0.500 L of a 4.00 M lead(II) nitrate solution to the beaker. Determine the mass of solid product formed (in grams).
When a solution is diluted by adding additional solvent, the concentration of solute changes hut the amount of solute present does not change. Explain. Suppose 250. mL of water is added to 125 mL of 0.55 1 M NaCl solution. Explain how you would calculate the concentration of the solution after dilution.
A student mixes four reagents together, thinking that the solutions will neutralize each other. The solutions mixed together are 50.0 mL of 0.100M hydrochloric acid, 100.0 mL of 0.200 M of nitric acid, 500.0 mL of 0.0100 M calcium hydroxide, and 200.0 mL of 0.100 M rubidium hydroxide. Did the acids and bases exactly neutralize each other? If not, calculate the concentration of excess H+ or OH ions left in solution.
Find the molarity of a 40.0% by mass aqueous solution of sulfuric acid, H2SO4, for which the density is 1.3057 g/mL.
Insulin is a hormone that controls the use of glucose in the body. How many moles of insulin are required to make up 28 mL of 0.0048 M insulin solution?
The units of parts per million (ppm) and parts per billion (ppb) are commonly used by environmental chemists. In general, 1 ppm means 1 part of solute for every 106 parts of solution. Mathematically, by mass: ppm=gsolutegsolution=mgsolutekgsolution In the case of very dilute aqueous solutions, a concentration of 1.0 ppm is equal to 1.0 g of solute per 1.0 mL, which equals 1.0 g solution. Parts per billion is defined in a similar fashion. Calculate the molarity of each of the following aqueous solutions. a. 5.0 ppb Hg in H2O b. 1.0 ppb CHCl3 in H2O c. 10.0 ppm As in H2O d. 0.10 ppm DDT (C14H9Cl5) in H2O
A solution is 0.1% by mass calcium chloride. Therefore, 100. g of the solution contains g of calcium chloride.
Calculate the number of moles and the mass of the solute in each of the following solutions: (a) 2.00 L of 18.5MH2SO4, concentrated sulfuric acid (b) 100.0 mL of 3.8105MNaCN, the minimum lethal concentration of sodium cyanide in blood serum (c) 5.50 L of 13.3 MH2CO, the formaldehyde used to fix tissue samples (d) 325 mL of 1.8106MFeSO4, the minimum concentration of iron sulfate detectable by taste in drinking water
Determine the molarity for each of the following solutions: (a) 0.444 mol of CoCl2 in 0.654 L of solution (b) 98.0 g of phosphoric acid, H3PO4, in 1.00 L of solution (c) 0.2074 g of calcium hydroxide, Ca(OH)2, in 40.00 mL of solution (d) 10.5 kg of Na2SO410H2O in 18.60 L of solution (e) 7.0103 mol of I2 in 100.0 mL of solution (f) 1.8104 mg of HCI in 0.075 L of solution
A common method for determining how much chloride ion is present in a sample is to precipitate the chloride from an aqueous solution of the sample with silver nitrate solution and then to weigh the silver chloride that results. The balanced net ionic reaction is :math>Ag+(aq)+Cl(aq)AgCl(s) ppose a 5.45-g sample of pure sodium chloride is dissolved in water and is then treated with a solution containing 1.15 g of silver nitrate. Will this quantity of silver nitrate be capable of precipitating all the chloride ion from the sodium chloride sample?
What is the difference between a solute and a solvent?