Principles of Organic Synthesis, Third Edition (Norman, Richard O.C.; Coxon, James M.)

Principles of Organic Synthesis, Third Edition (Norman, Richard O.C.; Coxon, James M.) Shelton Bank State University of New York at Albany, Albany, NY 12222 J. Chem. Educ., 1995, 72 (7), p A152 DOI: 10.1021/ed072pA152.1 Publication Date: July 1995 For  View and Download Click Her...

Read More

Organic Chemistry, 2 edition by Jonathan Clayden,

Organic Chemistry, 2 edition by Jonathan Clayden, Nick Greeves and Stuart Warren Topics reaction, chapter, reactions, bond, acid, carbonyl, group, compounds, carbon, bonds, carbonyl group, double bond, leaving group, conjugate addition, lone pair, carbon atom, nmr spectrum, starting material, interactive mechanism, carbonyl compounds Click...

Read More

Converting between "ppm" and molarity

Converting between "ppm" and molarity The definition of parts per million: 1 g solute per 1,000,000 g solution Now, divide both values by 1000 to get a new definition for ppm: ppm = 0.001 g per 1,000 g solution or: ppm = 1 mg solute per 1 kg solution Then, for an aqueous solution: ppm = 1 mg solute per liter of solution This last one works because the solution concentration is so low that we can assume the solution density to be 1.00 g/mL. Also, it's this last modification of ppm (the mg/L one) that allows us to go to molarity (which has...

Read More

What does "ppm" mean?

What does "ppm" mean?                                                         Go To Converting between "ppm" and molarity                                                                         The expression "1 ppm" means a given...

Read More

Molarity Problems

Molarity Problems                                                                                   Molarity Difination The equations I will use are: M = moles of solute / liters of solution and MV = grams / molar mass <--- The volume here MUST be in liters. Typically, the solution is for the molarity (M). However, sometimes it is...

Read More

Molarity Dafination and Explanation

Molarity Dafination and Explanation                                                                                 Molarity Problems As should be clear from its name, molarity involves moles. Boy, does it! The molarity of a solution is calculated...

Read More

Molality Problems

Molality Problems Problem #1: A solution of H2SO4 with a molal concentration of 8.010 m has a density of 1.354 g/mL. What is the molar concentration of this solution? Solution: 8.010 m means 8.010 mol / 1 kg of solvent8.010 mol times 98.0768 g/mol = 785.6 g of solute 785.6 g + 1000 g = 1785.6 g total for solute and solvent in the 8.010 m solution. 1785.6 g divided by 1.354 g/mL = 1318.76 mL 8.01 moles / 1.31876 L = 6.0739 M 6.074 M (to four sig figs) Problem #2: A sulfuric acid solution containing 571.4 g of H2SO4 per...

Read More