# How to measure concentration

by lil_agelu4life
Tags: concentration, measure
 P: 398 Even if you do not calculate the concentration in this experiement it could be interesting for you to find out. Here are some good links for the concept of concentation: http://en.wikipedia.org/wiki/Concentration http://chemistry.about.com/library/weekly/aa081003a.htm http://www.google.se/search?hl=sv&q=...%B6kning&meta= Concentration can be calculated via two forumlae, depending on if you are diluting a substance or if you want to find the concentration of eg. a certain amount (in mole) of a substance dissolved in a certain amount of solvent. Don't hesitate to ask if there is something you do not understand completely A basic knowledge of the concept of mole is good to have if one is dealing with concentration. $$n$$ = $$m/M$$ $$n$$ = number of mol $$m$$ = Amount of mass (in grams) $$M$$ = Relative molar mass / Relative Atomic mass (look at the O for Oxygen in a periodic table and you will see "16" (or "15.9994" depending on the accuracy of it). $$n$$ = $$V * C$$ $$V_1 * C_1$$ = $$V_2 * C_2$$ $$n$$ = number of mol $$V$$ = Volume in $$dm^3$$ ($$V_1$$ = Volume before, $$V_2$$ = Volume after) $$C$$ = Concentration in $$mol/dm^3$$ ($$C_1$$ = Concentration before, $$C_2$$ = Concentration after) Here are two examples of the usage of these formulae: I have 10 grams of table salt in 1 $$dm^3$$ water. If I increase the volume to 2 $$dm^3$$ what concentration will the final mixture have? $$m_(NaCl)$$ = 10 g $$V_1$$ = 1 $$dm^3$$ $$V_2$$ = 2 $$dm^3$$ $$M_(NaCl)$$ = 11 + 17 = 28 g/mol $$n_(NaCl)$$ = 10/28 mol $$C_(NaCl)$$ = (10/28)/1 = 10/28 $$V_1 * C_1$$ = $$V_2 * C_2$$ => $$C_2$$ = $$V_1 * C_1/V_2$$ = $$1 * (10/28) / 2$$ = 0.17....$$mol/dm^3$$