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If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. More Tutorials Join AUS-e-TUTE Contact Us Want chemistry games, drills, tests and more? You need to Join AUS-e-TUTE! Key Concepts ⚛ Concentration of a solution refers to the amount of solute dissolved in a given solvent to make a solution.
⚛ Concentration of a solution can be given in moles of solute per litre of solution (mol L-1 or mol/L or M), or, in moles of solute per cubic decimetre of solution (mol dm-3 or mol/dm3)(1)
⚛ Molarity is the term used to describe a concentration given in moles per litre. Alternative names for molarity are · amount of substance concentration (IUPAC preferred term) · amount concentration · molar concentration ⚛ Amount of substance concentration, molarity, has the units mol L-1 (or mol/L or M) or the equivalent SI units of mol dm-3 (mol/dm3) ⚛ Amount of substance concentration, molarity, the concentration of a solution in mol/L, mol L-1, mol dm-3 or mol/dm3, is given the symbol c (sometimes M). For a 0.01 mol L-1 HCl(aq) solution we can write : (i) [HCl(aq)] = 0.01 mol L-1 = 0.01 mol dm-3 = 0.01 M (ii) c(HCl(aq)) = 0.01 mol L-1 = 0.01 mol dm-3 = 0.01 M ⚛ Mathematical equation (formula or expression) to calculate the molarity of a solution (concentration in mol L-1) is c = n ÷ V c = concentration of solution in mol L-1 (mol/L or M) ⚛ This equation (formula or expression) can be re-arranged to find: (i) moles of solute given molarity and volume of solution: n = c × V (ii) volume of solution given moles of solute and molarity: V = n ÷ c Please do not block ads on this website. Molarity, amount of substance concentration, ConceptsAmount of substance concentration, molarity, is the term given for concentrations of solutions that are given in units of moles per litre (mol L-1 or mol/L or M) or, in SI units, moles per cubic decimetre (mol dm-3 or mol/dm3).
↪ Back to top Molarity Equation (amount of substance concentration equation)The amount of substance concentration, or the molarity of a solution, is given by the equation : c = amount concentration (molarity) in mol L-1 (or mol dm-3) Note that we often use square brackets around the formula of solvated species to indicate the molarity of a solution (the concentration of the solution in mol L-1): for example an aqueous solution of sodium chloride, NaCl(aq), with a molarity of 0.154 mol L-1 (or 0.154 mol dm-3) could be represented as c(NaCl(aq)) = 0.154 mol L-1 = 0.154 mol dm-3 [NaCl(aq)] = 0.154 mol L-1 = 0.154 mol dm-3 If you know the moles of solute in a solution, and, you know the volume of the solution, you can calculate the concentration of the solution in mol L-1 (or mol dm-3) using the mathematical equation c = n ÷ V But what if you know the concentration of the solution in mol L-1 (or mol dm-3) and the volume of the solution in L (or dm3), can you calculate how many moles of solute are present in the solution? You can even calculate the volume of the solution if you know how many moles of solute are present and its concentration in mol L-1 (or mol dm-3). In summary, to calculate (a) amount of substance concentration (molarity) in mol L-1 (or mol dm-3): c = n ÷ V (b) amount of solute in solution in mol : n = c × V (c) volume of solution in L (or dm3): V = n ÷ c ↪ Back to top Examples of Molarity Calculations with Worked SolutionsApply the following 5 steps to solve the molarity problems below:
(1) Calculating Amount of Substance Concentration, Molarity (c = n ÷ V)Question: Calculate the amount of substance concentration in mol L-1 (molarity) of an aqueous sodium chloride solution containing 0.125 moles sodium chloride in 0.50 litres of solution. Solution: Step 1: What is the question asking you to calculate? c(NaCl(aq)) = molarity (concentration in mol L-1) of solution = ? mol L-1 Step 2: What information has been given in the question? Extract the data from the question Step 3: What is the relationship between what you know and what you need to find? Write the equation: Step 4: Substitute the values into the equation for molarity and solve: [NaCl(aq)] = c(NaCl(aq)) = 0.125 mol ÷ 0.50 L = 0.25 mol L-1 (or 0.25 mol/L or 0.25 M) (Note: only 2 significant figures are justified) Step 6: Write the answer: [NaCl(aq)] = 0.25 mol L-1 (2) Calculating Amount of Solute (n = c × V)Question: Calculate the moles of copper sulfate in 250.00 mL of 0.020 mol L-1 copper sulfate solution, CuSO4(aq). Solution: Step 1: What is the question asking you to calculate? n(CuSO4) = moles of solute = ? mol Step 2: What information has been given in the question? Extract the data from the question: Step 3: What is the relationship between what you know and what you need to find? Write the equation: Step 4: Substitute the values into the molarity equation and solve: n(CuSO4) = 0.020 mol L-1 × 0.25000 L = 0.005000 mol = 0.0050 mol (Note: only 2 significant figures are justified) Step 5: Write the answer: n(CuSO4(aq)) = 0.0050 mol (3) Calculating Volume of Solution (V = n ÷ c)Question: Calculate the volume in litres of a 0.80 mol L-1 aqueous solution of potassium bromide containing 1.60 moles of potassium bromide. Solution: Step 1: What is the question asking you to calculate? V(KBr(aq)) = volume of solution in litres = ? L Step 2: What information has been given in the question? Extract the data from the question: Step 3: What is the relationship between what you know and what you need to find? Write the equation: Step 4: Substitute the values into the equation and solve: V(KBr(aq)) = 1.60 mol ÷ 0.80 mol L-1 = 2.00 L = 2.0 L (Note: only 2 significant figures are justified) Step 5: Write the answer: V(KBr(aq)) = 2.0 L ↪ Back to top Problem Solving: Amount of Substance Concentration (molarity)The Problem: Chris the Chemist has been given a 250.00 mL volumetric flask and asked to use it to make an aqueous solution of sodium chloride, NaCl, with a concentration of 0.100 mol L-1 for a corrosion experiment. The sodium chloride, NaCl, is available as an analytical reagent composed of white crystals. Determine the mass in grams of sodium chloride that Chris the Chemist will need to weigh out. Solving the Problem using the StoPGoPS model for problem solving:
↪ Back to top Sample Question: Mass Concentration A 25.00 mL aliquot of 0.0264 mol L-1 AgNO3(aq) is added to a 150.00 mL volumetric flask and distilled water is added until the meniscus sits on the mark when viewed at eye-level. ↪ Back to top
Footnotes: (1) There are many other ways to measure concentration of solutions in chemistry. Some common ones are listed in the introduction to solutions tutorial. (2) This is an important assumption because the volumetric flask has been calibrated to hold the stated volume at a given temperature. Volume is dependent on temperature. (3) This is an important assumption because if the water used to make the solution already contains dissolved sodium chloride we do not know how much sodium chloride is already present. Any dissolved sodium chloride in the water will increase the concentration of sodium chloride in the final solution. (4) This is an important assumption because if the reagent contains impurities, some of the mass we weigh out will NOT be due to NaCl and the concentration of NaCl will actually be less than we think. ↪ Back to top |