MCAT Non-Sim General Chemistry GCHM - Stoichiometry 5 years Question Information Q-Bank MCAT Sim Non-Sim Subject General Chemistry Foundation GCHM - Stoichiometry Validity 5 years Author(s) Reyes, V. M. Reviewer(s) 0000000 Editor(s) 0000000 Passage Media Media ID(s) Passage How many molecules are there in a gram of benzene? How many Na+ and Cl- ions are there in a gram of table salt? And how many molecules are there in a liter of water? The above questions require one to be able to relate an atomic or molecular property (number of atoms or molecules) of an entity (the element or compound in question) to a macroscopic property (mass in grams) of the same entity. Even before the middle of 1800’s, the total charge of a mole of electrons has been been known from electrolysis experiments wherein a certain mass, say 0.10 gram, of a metal, say silver, from a solution of silver nitrate has been deposited onto the

6.0221 x 1023 per mole or 6.0221 x 1023 mole -1 Avogadro’s Number: anode after passage of an electric current , say 1.0 amperes, via electrolysis, for a certain amount of time, say 1.5 min. Here, the number of coulombs used is 90 amp.-sec. or 90 coul. Since silver went from Ag+ in AgNO3 to Ag0 in Ag metal, it went from an oxidation state of +1 to 0, that is, for every atom of Ag+, one electron (e-) was consumed to convert it to Ag (metal), which is the same as saying that for every mole of Ag+, one mole of e- was consumed. From the periodic table, the molar mass of Ag is 107.87 gm./mol., thus the 0.10 gm. of Ag deposited represents 9.27 x10-4 mol., which also represents the number of moles of electrons consumed. Since we want to find the charge of one mole of e-, we would divide 90 coul. found above, by 9.27 x10-4 mol. of e- , and the result is 97,090 coul./mol. The exact number is actually 96,485 Coul./mol. for Faraday’s constant, the charge of a mole of electrons. Now, to determine the value of Avogadro’s number, we need to find the charge of a single e-. In the early 1900s, Robert Millikan came up with his famous oil-drop experiment which showed convincingly that this charge is 1.602 x 10-19 Coul. Dividing Faraday’s constant with this value, we obtain 6.023 x 1023 /mol. for Avogadro’s Number. The exact number is shown in Figure 1. Figure 1 Avogadro’s Number: 6.0221 x 1023 per mole or 6.0221 x 1023 mole -1 Avogadro’s number is the number of particles (atoms or molecules) in a mole of that substance or compound. Most calculations involving Avogadro’s number would Involve the molar mass of the substance in question, which is the atomic mass (or atomic weight) in grams. Modern and more accurate methods of determining Avogadro’s number make use of x-ray crystallographic techniques using the unit cell dimensions of the crystal lattice and the molar volume of the compound constituting the crystal.

L. Pauling, General Chemistry Passage References PMID/Book Title of Publication or Book 000000000 L. Pauling, General Chemistry (N/A) en.wikipedia.com, youtube.com (N/A) Author’s own lecture notes. Question Attributes #1 Topic Blueprint Avogadro’s Number Competency MCAT: BS-2: Application of Concepts & Principles To test student’s understanding of the mole concept and how to apply it to chemical calculations using Avogadro’s number . Objective Media ID(s) 00000000 Question ID 00000000 Question Stem #1 Calculate the number of Ca++ cations and Cl- anions in a pound of deicing salt (CaCl2) whose purity is 98.5% by weight. Given: 1 lb. = 0.4536 kg.; atomic weights of Ca and Cl are 40.078 and 35.453, respectively. Answer Choices #1 A) 2.424 x 1024 of Ca++ cations and 1.212 x 1024 Cl- anions B) 2.424 x 1024 of Ca++ cations and 4.848 x 1024 Cl- anions C) 2.461 x 1024 of Ca++ cations and 4.922 x 1024 Cl- anions D) 3.562 x 1024 of Ca++ cations and 7.124 x 1024 Cl- anions Correct: B)

The correct answer is B. The solution goes as follows: Explanation #1 The correct answer is B. The solution goes as follows: 1.0 lb. = 453.6 grams, but at 98.5% purity, this goes down to 446.8 gm. Molecular weight of CaCl2 is equal to 40.078 + 2(35.453) = 110.98 gm./mol. Number of moles of CaCl2 in 446.8 gm. Is 446.8 gm./(110.98 gm./mol.) or 4.026 moles. Thus there are 4.026(6.0221 x 1023) = 24.24 x 1023 molecules or 2.424 x 1024 molecules of CaCl2 . Therefore there are 2.424 x 1024 cations of Ca++ and twice that, i.e., 4.848 x 1024 anions of Cl- in the sample. (Choice A) Here, the number of Cl- anions was halved instead of doubled. (Choice C) Here, the purity was not taken into account; the sample was taken to be 100% pure instead of only 98.5% pure. (Choice D) Here, the MW of CaCl2 was calculated by mistake to be 40.078 + 35.453 = 75.531 (i.e., the atomic weight of Cl was not doubled. Educational objective: To test student’s understanding of the mole concept and how to apply it in chemical calculations using Avogadro’s number .

0000000 0000000 References #1 PMID/Book Title of Publication or Book 000000000 000000000 0000000 Verifications #1 Yes / No The question is at the Application or higher cognitive level. Yes / No The question is based on a realistic clinical scenario. Yes / No The question has at least one close distracter, and other options have educational value. Yes / No The question is appropriate to the entry level of nursing practice. Yes / No The explanation is short and concise, yet thorough. Yes / No The question has an appropriate table/flow chart/illustration.