Concentration Expressed in molarity, e.g.: 2 M NaCl = 2 “molar” NaCl = 2 moles NaCl per liter of solution (where 1 mole = 6.023 x 1023 molecules)

How to make a 2M NaCl stock solution You will have to put 2 moles of NaCl inside 1 liter of solution From the Periodic Table, 1 mole of NaCl weighs: 1 x Na = 22.99 grams, 1 x Cl = 35.45 grams 1 x NaCl = 22.99+35.45 = 58.44 grams So the formula weight of NaCl = 58.44 grams In a beaker, slowly dissolve 2 x 58.44 = 116.88 grams NaCl into about 800 ml water, then pour into graduated cylinder and make volume up to exactly 1 liter (=1000 ml)

Algebra Reminder Whatever you do on one side of the equal sign, you must also do on the other side of the equal sign: = = + + = +

How to make 200 ml of a 2 M NaCl stock solution 2 M NaCl = 116.88 g / liter = 116.88 g /1000 ml But we don’t want 1000 ml, we want 200 ml! Use algebra: 116.88 g = x g 1000 ml 200 ml Now multiply by 200 ml on both sides to find x: (116.88 g)(200 ml) = (x g)(200 ml) = x = 23.38 g (1000 ml) (200 ml) (So you will add 23.38g NaCl into a total of 200ml)

How to use stock solutions How would you make 150 ml of a 1M NaCl solution from a 3M NaCl stock solution? In other words, how many ml of your stock solution must go into 150 ml? C1V1 = C2V2 (x ml)(3M) = (150 ml)(1M) Now divide both sides by 3M to find x: (x ml)(3M) = x ml = (150 ml)(1M) = 50 ml (3M) (3M)

Fold Dilutions Most common notation, e.g. for a 10-fold dilution: 1 : 10 = 1 part (stock solution) into a total of 10 parts = 1 part (stock solution) + 9 parts (water)

A B C Serial Dilutions Bob's.ppt transferring 1 ml 1 ml A = concentrated stock solution +9 ml +9 ml B = 1:10 dilution of A C = 1:10 dilution of B and a 1:100 dilution of A Bob's.ppt http://www.geocities.com/bkip20002/index.html Graphics by Bob http://home.att.net/~kip20002/ A B C