Bellwork Friday Draw and finish filling in the mole road map to the right.

Dilute solution- Contains a small amount of solute 16. 2 Concentration -a measure of the amount of solute per given amount of solvent or solution. Measured in units of molarity Dilute solution- Contains a small amount of solute Concentrated solution- Contains a large amount of solute

Molarity(M) = moles of solute liters of solution A 1 M solution contains 1 mole of solute per 1 L of solution. A 0.5 M NaCl solution has 0.5 mol NaCl in 1 L total solution.

EXAMPLES: What is the concentration in molarity of a solution made with 1.25 mol NaOH in 4.0 L of solution? # mol # L M = M = ? #mol = 1.25 mol V = 4.0 L M = 1.25 mol 4.0 L = 0.3125 M = 0.31 M

Question A student measures 58.5g of NaCl and dissolves in into a 1.0 L of water. What is the molarity of the solution?

Dilutions Diluting a solution won’t change the number of moles of solute. It only changes the concentration. M1V1 = M2V2 This formula works only for dilution of a solution!!!! Don’t use it for a reaction!

How would you make 500. mL of 6.0 M NaOH from 12.0M NaOH? To make the solution, you would measure out 250. ml of the concentrated base and add it to water until you have 500. ml of solution. M1 = 12.0M V1 = ? M2 = 6.0M V2 = 500. mL (12.0M)(x) = (6.0M)(500. ml) X = 250. ml How many grams of NaOH would be necessary to make 250.ml of 12.0 M solution? 250. ml NaOH 1L NaOH 12.0 mol NaOH 40.0 g NaOH = 1000 ml 1 L NaOH 1 mol NaOH 120. g NaOH

Question A solution is made with 10.5g AgCl dissolved in 2.0L of water. What is the molarity of the silver chloride solution?

EXAMPLE: For a lab we need 500. mL of 4.25 M HCl. Concentrated HCl is 12.0 M. How do we make up the solution? M1V1 = M2V2 M2 = 4.25 M V2 = 500. mL M1 = 12.0 M V1 = ? (12.0 M)(V1) = (4.25M)(500. mL) V1 = 177 mL Add 177 ml of concentrated hydrochloric acid to enough water to make 500 ml of solution.

Solution Stoichiometry use Molarity as a conversion factor# # mol 1 L 1 L # mol

EXAMPLES: What volume of 0.246 M nitric acid is required to react completely with 0.386 L of 0.0515 M Ba(OH)2? 2HNO3 + Ba(OH)2 → Ba(NO3)2 + 2H2O 0.386 L Ba(OH)2 0.0515 mol Ba(OH)2 2 mol HNO3 1 L HNO3 1 L Ba(OH)2 1 mol Ba(OH)2 0.246 mol HNO3 molarity mol-mol ratio molarity = 0.162 L HNO3

Calculate the number of grams of carbon dioxide that can react with 0 Calculate the number of grams of carbon dioxide that can react with 0.135 L of a 0.357 M solution of potassium hydroxide, according to the following reaction: 2KOH + CO2  K2CO3 + H2O 0.135 L KOH 0.357 mol KOH 1 mol CO2 44.0 g CO2 1 L KOH 2 mol KOH 1 mol CO2 = 1.06 g CO2 molarity mol-mol ratio GFM

% Solutions Percent by volume % v/v = volume solute (mL) x 100 volume solution (mL) Percent mass/volume % m/v = mass solute(g) x 100

EXAMPLES: Rubbing alcohol is 70.0% isopropyl alcohol by volume. How many mL of isopropyl alcohol are in a 250 mL bottle of this solution? % v/v = volume solute (mL) x 100 volume solution (mL) % v/v = 70.0 % v solution = 250 mL v solute = X 70.0 % = __X__ x 100 250 mL X = 175 mL

If the % mass/volume (m/v) for a solute is 6 If the % mass/volume (m/v) for a solute is 6.80 %, and the volume of the solution is 42.6 mL, what mass of the salt is present? % m/v = mass solute (g) x 100 volume solution (mL) % m/v = 6.80 % v solution = 42.6 mL g solute = X 6.80 % = __X__ x 100 42.6 mL X = 2.8968 g = 2.90 g

Warm-up Tuesday Calculate the number of grams of carbon dioxide that can react with 0.135 L of a 0.357 M solution of potassium hydroxide, according to the following reaction: 2KOH + CO2  K2CO3 + H2O