SolutionstoMiLab2prob - x = 0.4(255ul)/5 = 20.4 ul of 5M...

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Solutions to Practice Problems, Lab Manual p. M-4 (Intro to Mol Biology): I tend to set up these problems as an algebra equation, to make sure I account for everything: (Conc initial )(volume initial ) = Conc final )(Volume final ) -then solve for the unknown. For the first two questions below, I show this approach first (a), followed by the way the lab manual does (ie calculate your dilution factor first)(b). Don’t forget to keep track of your units! The answer is wrong without them, and it helps you double check that you set up the problem correctly (ie if the units don’t work out, something is off). 1) CaCl 2 solution: a) (Conc initial )(volume initial ) = Conc final )(Volume final )- you have what you want for final, and your starting concentration, so you solve for how much initial volume you need to take: (5M)(volume initial ) = (0.4M)(255ul)
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Unformatted text preview: x = 0.4(255ul)/5 = 20.4 ul of 5M stock CaCl 2 255ul-20.4ul = 234.6ul water b) dilution in conc = .4M/5M = .08 x final volume 255ul = 20.4ul water is the same as in a) 2) NaAcetate question: a) The set up is the same to start, but you are solving for final volume this time: 50ul (8M) = 0.025M (X) solve for X gives X = 50ul (8M)/0.025M = 16000ul Converting microliters to ml: 16,000 ul (1 ml / 10 3 ul) = 16 ml b) You want to make X ul of 0.025 M stock. Therefore the dilution factor is 0.025 / 8 M = 0.003125. (0.003125) (X) = 50 ul Solving for X gives X=50 ul/ 0.003125 = 16,000 ul 3) NaCl question: 58.4 g/mol, thus a 1 Molar solution: 1M = 58.4g/1L, divide both parts by 2 to get 0.5L = 29.2g/.5L, so 29.2 g is needed to make 500ml of 1M. To get a 0.2 M solution, multiply by .2 , (0.2) 29.2g = 5.84g...
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