Here is a helping hand with some of the new material on Molarity: Remember: Not all substances that we use in chemical reactions are in the same PHASE of matter Solids (s) Liquids (l) Gases (g) And then to complicate matter (that’s a joke) – Liquids that are a uniform mix of solute and solvent are aqueous solutions (aq)…they are made from ions. It is easier for many reactions to take place in a liquid because they can move more easily in liquids…. That’s another chapter in itself!

Molarity = Number of moles of solute liters of solution What is the molarity of 40.0 grams of calcium carbonate dissolved in 1.0 L of water? What if it is dissolved in 2.0 L of water? What if 20.0 grams of the rock salt are dissolved in 1.0 L of water? …10.0 grams in 500mL? …40.0 grams in 250 mL?

A saline (salt) contains 0. 90 g of NaCl in 100 mL of solution A saline (salt) contains 0.90 g of NaCl in 100 mL of solution. Calculate M: .90 g ( 1 mole) 58.5 g = .0154 moles NaCl .0154 moles .100 L = .154 M

A salt solution has a volume of 250 mL and contains 0. 70 mole of NaCl A salt solution has a volume of 250 mL and contains 0.70 mole of NaCl. What is the molarity? 0.70 moles of NaCl = .250 L 2.8M

Now, how many grams is this? How many moles of silver nitrate are needed to make 350 mL of a 2.0 M AgNO3 solution? X moles .350L = 2.0M So, .350 x 2.0 = .70 moles Now, how many grams is this? There are 170 g in a mole of silver nitrate… .70 moles x 170 g = 119 = 120 g Ag NO3 1 mole

400.0 mL of a .40 M solution of BaCl2 First let’s find the amount of moles of barium chloride that one must gather from a container of the salt. Now convert that amount of moles to grams: .

e. A solution that is made from dissolving 17 e. A solution that is made from dissolving 17.00 g of NaNO3 to make 160.0 mL… Would have a molarity calculated as follows: Molarity is moles liter … So we must find moles, Then plug that in and,

f. 35.5 g of solid Na2SO4 can make how many liters of .25 M solution? X mole = M ? Liters 1 Cross multiply and divide = .moles x 1 / (.25) = 1 liter Well, how many moles is 35.5 g of sodium sulfate? Call that X moles How many Liters would .25 moles dissolve in to make a .25 M solution? .

How many grams of Silver will be produced from 85. 0 mL of 1 How many grams of Silver will be produced from 85.0 mL of 1.50 M silver nitrate and copper metal? For this problem you will have to begin with a balanced equation: __AgNO3 + __Cu  Now you need to add the phase symbols and coefficients for the mole to mole ratio. X moles = 1.50 M .085 L,

Now we need to determine the mole to mole ratio of the given and find: The balanced equation shows us that for every 2 moles of silver nitrate reacted with 1 mole of Cu that 2 moles of Ag will be produced along with 1 mole of copper nitrate. (It is important to note that since the Cu metal is ‘in excess’ that all the silver nitrate gets used up and the full 2 moles of product are produced!) 2 Ag NO3 (aq) + 1 Cu (s)  2 Ag (s) + 1 Cu(NO3)2 (aq)

h. Calculate the # of Barium Sulfate grams when 125mL of 1 h. Calculate the # of Barium Sulfate grams when 125mL of 1.20 M Barium nitrate reacts with excess sulfuric acid. Write the equation: __Ba(NO3)2 +__H2SO4 Now balance it. Find the moles of the given: x moles = 1.20 M ,

Now determine the balanced molar ratio for all the other reactants and products. 1 Ba(NO3)2 + 1H2SO4  1BaSO4 + 2HNO3 Find the Barium Sulfate grams: