Dilution A chemist makes 220. mL of aluminum chloride (AICI3) working solution by adding distilled water to 90.0 mL of a 3.03 M stock solution of aluminum chloride in water. Calculate the concentration of the chemist's working solution. Be sure your answer has the correct number of significant digits. M

Dilution A chemist makes 220. mL of aluminum chloride (AICI3) working solution by adding distilled water to 90.0 mL of a 3.03 M stock solution of aluminum chloride in water. Calculate the concentration of the chemist's working solution. Be sure your answer has the correct number of significant digits. M

Introductory Chemistry: An Active Learning Approach

6th Edition

ISBN:9781305079250

Author:Mark S. Cracolice, Ed Peters

Publisher:Mark S. Cracolice, Ed Peters

Chapter16: Solutions

Section: Chapter Questions

Problem 16.1TC

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Similar questions

What mass of solid NaOH (97.0% NaOH by mass) is required to prepare 1.00 L of a 10.0% solution of NaOH by mass? The density of the 10.0% solution is 1.109 g/mL.
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You wish to prepare 1 L of a 0.02-M potassium iodate solution. You require that the final concentration be within 1% of 0.02 M and that the concentration must be known accurately to the fourth decimal place. How would you prepare this solution? Specify the glassware you would use, the accuracy needed for the balance, and the ranges of acceptable masses of KIO3 that can be used.
A soft drink contains an unknown mass of citric acid, C3H5O(COOH)3. It requires 6.42 mL of 9.580 × 10−2-M NaOH to neutralize the citric acid in 10.0 mL of the soft drink. C3H5O(COOH)3(aq) + 3 NaOH(aq) → Na3C3H5O(COO)3(aq) + 3 H2O(ℓ) Determine which step in these calculations for the mass of citric acid in 1 mL soft drink is incorrect? Why? n (NaOH) = (6.42 mL)(1L/1000 mL)(9.580 × 10−2 mol/L) n (citric acid) = (6.15 × 10−4 mol NaOH) × (3 mol citric acid/1 mol NaOH) m (citric acid in sample) = (1.85 × 10−3 mol citric acid) × (192.12 g/mol citric acid) m (citric acid in 1 mL soft drink) = (0.354 g citric acid)/(10 mL soft drink) Determine the correct result.
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
On Easter Sunday, April 3, 1983, nitric acid spilled from a tank car near downtown Denver, Colorado. The spill was neutralized with sodium carbonate: 2HNO3(aq)+Na2CO3(aq)2NaNO3(aq)+H2O(l)+CO2(g) a. Calculate H for this reaction. Approximately 2.0 104 gal nitric acid was spilled. Assume that the acid was an aqueous solution containing 70.0% HNO3 by mass with a density of 1.42 glcm3. What mass of sodium carbonate was required for complete neutralization of the spill, and what quantity of heat was evolved? (Hf for NaNO3(aq) = 467 kJ/mol) b. According to The Denver Post for April 4, 1983, authorities feared that dangerous air pollution might occur during the neutralization. Considering the magnitude of H, what was their major concern?