# Molarity 2. Molarity (M) this is the most common expression of concentration M = molarity = moles of solute = mol liters of solution L Units are.

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Molarity 2. Molarity (M) this is the most common expression of concentration M = molarity = moles of solute = mol liters of solution L Units are mol L-1 or mol/L (moles per liter) A solution that is 1.0 molar (1.0M ) contains 1.0 mol of solute per liter of solution M = molarity (mol L-1) n = number of moles (mol) V = volume (L) n M V M = n/V (mol L-1) n = M  V (mol) V = n /M (L)

Example 1: Calculate the molarity of a solution made by dissolving 11.5g of solid NaOH in enough water to make 1.50 L of solution. Step 1: Find the # of moles of NaOH n = m = g MM g mol-1 = mol Step2: Find the molarity M olarity = moles of solute = 0.288mol liters of solution 1.50 L = M NaOH

Example 2: Calculate the molarity of a solution made by dissolving 1.56g of gaseous HCl in enough water to make 26.8 mL of solution. Step 1: Find the # of moles of HCl n = m = g MM g mol-1 = mol HCl = 4.27 10-2 mol HCl Step2: Convert milliliters to Liters 26.8mL/1000L = L = 2.68  10-2 L Step3: Find the molarity M olarity = moles of solute = 4.27 10-2 mol liters of solution  10-2 L = 1.59 M HCl

Example 3: Give the concentrations of all the ions in each of the following solutions:
a M Co(NO3)2 b. 1 M FeCl3 a. Co(NO3)2 (s)  Co2+ (aq) + 2NO3- (aq) 1 mol Co(NO3)2 (s)  1 mol Co2+ (aq) + 2 mol NO3- (aq) Therefore a solution that is 0.5 mol Co(NO3)2 contains 0.50 M Co2+ and 1 M NO3- b. FeCl3 (s)  Fe3+ (aq) + 3Cl- (aq) 1 mol FeCl3 (s)  1 mol Fe3+ (aq) + 3 mol Cl- (aq) Therefore a solution that is 1 M FeCl3 contains 1 M Fe3+ and 3 M Cl-

Example 4: How many moles of Ag+ ions are present in 25mL of a 0
Example 4: How many moles of Ag+ ions are present in 25mL of a 0.75 M AgNO3 solution? Molarity of solution = 0.75 M Volume of solution = 25mL moles of Ag+ ions = ? A 0.75 M AgNO3 solution contains 0.75 M Ag+ ions and 0.75 M NO3- ions Convert mL to L 25mL / 1000L = 2.5  10-2 L Molarity = n or n = Molarity  Volume V (2.5  10-2 L)  (0.75 M Ag+) = 1.9  10-2 mol Ag+

Calculating mass from Molarity
A chemist is analyzing the alcohol content of a wine from Napa. To do this she needs to use 1.00L of an M K2Cr2O7 solution. (molar mass of K2Cr2O7 = g mol-1) How much solid K2Cr2O7 must be weighted out to make this solution? Molarity of solution = M Volume of solution = 1.00 L grams of K2Cr2O7 = ? Step 1: Calculate the moles of K2Cr2O7 present. Molarity = n or n = Molarity  Volume V = (1.00 L)  (0.200 M K2Cr2O7) = mol K2Cr2O7

Step 2: Convert moles into grams
n = mass or mass = Molar mass  n molar mass = (0.200 mol K2Cr2O7)  (294.2g K2Cr3O7) = 58.8g K2Cr2O7

Making a Standard Solution
A standard solution is a solution whose concentration is accurately known. This done by weighing out a sample of solute, transferring it completely to a volumetric flask, and adding enough solvent to bring the volume up to the mark on the neck of the flask.

Formulae for solution problems
n = number of moles (mol) m = mass (g) MM = molar mass (g mol-1) M = Molarity (mol L-1) n = number of moles (mol) V = volume (L)

Dilution To save space in out Prep room we buy solutions in concentrated form, i.e. 18M HCl (18 mol L-1). We call these stock solutions. The process of adding more solvent to a solution is called dilution. A typical dilution involves determining how much water must be added to an amount of stock solution to achieve a solution of the desired concentration. The amount of solute after dilution = moles of solute before dilution Remains constant M = moles of solute volume (L) Decreases Increases (water added)

Example: We want to prepare 500. mL of 1
Example: We want to prepare 500. mL of 1.00 M acetic acid, CH3COOH, from a 17.5 M stock solution of acetic acid. What volume of the stock solution is required? (b) water is added to dilute it (c) final diluted solution has a volume of 0.5L and a molarity of 1 M acetic acid (a) What volume of 17.5 M acetic acid do you have to remove?

Example: We want to prepare 500. mL of 1
Example: We want to prepare 500. mL of 1.00 M acetic acid, CH3COOH, from a 17.5 M stock solution of acetic acid. What volume of the stock solution is required? Step1: Find the number of moles of acetic acid needed in final solution n = molarity of dilute solution  volume of dilute solution n = 1.00 molL-1  0.5 L n = mol CH3COOH Step2: Find the volume of 17.5 M acetic acid that contain mol of CH3COOH. We will call this unknown volume V. V= moles of solute = mol CH3COOH Molarity mol L-1 V = L or 28.6 mL

(a) 28.6 mL of 17.5 M acetic acid solution is transferred to a volumetric flask that already some water. (b) water is added to the flask up to the 500 mL mark. (c) The final solution is 1.00 M acetic acid

M1  V1 = moles of solute = M2  V2
Because the moles of solute remain the same before and after dilution, we can write Final Conditions Initial Conditions M1  V1 = moles of solute = M2  V2 Volume before dilution Volume after dilution Molarity before dilution Molarity after dilution 17.5 M  L = moles of solute = 1.0 M  0.5 L 17.5 M  L = moles of solute = mol 1.0 M  0.5 L = moles of solute = mol M1  V1 = moles of solute = M2  V2

Use this equation to solve this problem: M1  V1 = M2  V2
What volume of 16 M sulfuric acid must be used to prepare 1.5 L of a 0.10 M H2SO4 solution? M1 = 16 molL-1 M2 = 0.10 molL-1 V1 = ? V2 = 1.5 L = 9.4  10-3 L = 9.4 mL

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