CHEM 1415 1st EditionLecture 2Unit: Solution StoichiometryOutline of Current Lecture I. Examples for testa. Example 1: finding the number of moles in the moleculeb. Example 2: finding the molarityc. Example 3: finding moles with volume givend. Example 4: finding moles with volume givenCurrent LectureExample 1:Finding the number of moles in the moleculeS8Molarity=8∗32=256gmol(18.3 g)(1 mol256 g)=0.04 molExample 2:20 g of NaOH in 100 mL of solutionWhat is the Molarity?Na→ 1∗23=23(100 mL)(1 L1000 mL) = 0.1 LO→ 1∗16=16Number of moles=(20 g)(1 mol40 g)=0.5 molH → 1∗1=1molarity=0.5 mol0.1 L=5molL=5 MMolar mass = 40 g/molExample 3:How many moles of HBr are in a 40mL solution of water with a molarity of 1.5 M?Moles of HBr=(Molarity) (Liters of solution)¿(1.5 M) (40 mL)(1 L1000 mL)¿0.06 molThese notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.Example 4:How many moles of NaClO are in a solution of 750.0 mL with a molarity of 0.806 M?molarity=molesof soluteLitersof solutionmolesof solute=(molarity) (Liters of solution)¿(0.806mo lL)(750 mL)(1 L1000mL)¿0.605
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