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BB lecture 9-08-08 carbon cpds

# BB lecture 9-08-08 carbon cpds - Learning objectives...

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Learning objectives Chapter 3 continued – Know how to calculate the concentrations of solutions percentage, molarity, and osmolarity. Be able to convert between expressions. Chapter 4 (pp.58-66) Carbon compounds Explain the basis for carbon’s ability to form structurally diverse molecules. Define organic chemistry. Be able to draw examples of different carbon “skeletons”. Distinguish among the three types of isomers: structural, geometric, and enantiomer (aka optical). Be able to recognize, name and draw major functional groups. Identify types of organic compounds by the ending of the name. Apply your understanding of polarity to infer solubility characteristics of organic compounds.

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Solutions – consist of a solute dissolved in a solvent. In living cells, the solvent is water. Solutions may be expressed in different ways – for example: - Percent - Molarity A 5% glucose solution contains 5 g of glucose in how many milliliters (ml) of solution? Another method of expression is molarity , or moles of solute per liter soln. A mole of a compound (cpd.) is how many grams of that cpd. is equal to its Mol. Mass. (aka Mol. Wt….MW, and also as Formula Wt., ‘FW’) Example: What is the mol. mass of glucose [ C 6 H 12 O 6 ]? atomic masses are C = 12, H = 1, O = 16 A 1-molar ( 1 M ) solution of glucose contains ___ g glucose (1 mole) dissolved in enough water to make 1 liter of solution.
How much glucose would be needed to make 100 ml of a 0.1M solution? [M. Mass of glucose = 180 g] Answer: It would take 18 g in a liter to make a 0.1M solution. But, a volume of 100 ml would only require 1/10 th of 18, or 1.8 g.

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The molecular mass of glucose (C 6 H 12 O 6 ) is 180 g. To make a 0.5 M solution of glucose, you should do which of the following? A) Dissolve 180 g of glucose in a small volume of water, and then add more water until the total volume of solution is 0.5 L.
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BB lecture 9-08-08 carbon cpds - Learning objectives...

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