Solutions Molarity ICS 1 Molarity M A concentration

Molarity (M) A concentration that expresses the moles of solute in 1 L of

Units of Molarity 2. 0 M HCl = 2. 0 moles HCl 1 L

Molarity Calculation Na. OH is used to open blocked sinks, to treat cellulose in

Calculating Molarity 1) 4. 0 g Na. OH x 1 mole Na. OH =

Test question 1 A KOH solution with a volume of 400 m. L contains

Answer ICS an Dr 2) 5 M M = 2 mole KOH = 5

Test question 2 A glucose solution with a volume of 2. 0 L contains

Answer 2 A glucose solution with a volume of 2. 0 L contains 72

Molarity Conversion Factors A solution is a 3. 0 M Na. OH. . Write

Test question 3 Stomach acid is a 0. 10 M HCl solution. How many

Answer 3 3) 1500 m. L x 1 L = 1000 m. L 1.

Test question 4 How many grams of KCl are present in 2. 5 L

Answer 4 3) 2. 5 L x 0. 50 mole x 74. 6 g

Test question 5 How many milliliters of stomach acid, which is 0. 10 M

Answer 5 2) 0. 15 mole HCl x 1 L soln x 1000 m.

Test question 6 How many grams of Na. OH are required to prepare 400.

Answer 6 2) 400. m. L x 1 L = 0. 400 L 1000

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Solutions Molarity ICS 1

Molarity (M) A concentration that expresses the moles of solute in 1 L of solution Molarity (M) = moles of solute 1 litre solution ICS 2

Units of Molarity 2. 0 M HCl = 2. 0 moles HCl 1 L HCl solution 6. 0 M HCl = 6. 0 moles HCl 1 L HCl solution ICS 3

Molarity Calculation Na. OH is used to open blocked sinks, to treat cellulose in the making of nylon, and to remove potato peel commercially. If 4. 0 g Na. OH are used to make 500. m. L of Na. OH solution, what is the molarity (M) of the solution? ICS 4

Calculating Molarity 1) 4. 0 g Na. OH x 1 mole Na. OH = 0. 10 mole Na. OH 40. 0 g Na. OH 2) 500. m. L x 1 L_ 1000 m. L 3. 0. 10 mole Na. OH 0. 500 L = 0. 20 mole Na. OH 1 L = 0. 20 M Na. OH ICS 5

Test question 1 A KOH solution with a volume of 400 m. L contains 2 mole KOH. What is the molarity of the solution? Dr an o 1) 8 M 2) 5 M 3) 2 M ICS 6

Answer ICS an Dr 2) 5 M M = 2 mole KOH = 5 M 0. 4 L o A KOH solution with a volume of 400 m. L contains 2 moles of KOH. What is the molarity of the solution? 7

Test question 2 A glucose solution with a volume of 2. 0 L contains 72 g glucose (C 6 H 12 O 6). If glucose has a molar mass of 180. g/mole, what is the molarity of the glucose solution? 1) 0. 20 M 2) 5. 0 M 3) 36 M ICS 8

Answer 2 A glucose solution with a volume of 2. 0 L contains 72 g glucose (C 6 H 12 O 6). If glucose has a molar mass of 180. g/mole, what is the molarity of the glucose solution? 1) 72 g x 1 mole x 180. g ICS 1 = 2. 0 L 0. 20 M 9

Molarity Conversion Factors A solution is a 3. 0 M Na. OH. . Write the molarity in the form of conversion factors. 3. 0 moles Na. OH and 1 L Na. OH soln ICS 1 L Na. OH soln 3. 0 moles Na. OH 10

Test question 3 Stomach acid is a 0. 10 M HCl solution. How many moles of HCl are in 1500 m. L of stomach acid solution? 1) 15 moles HCl 2) 1. 5 moles HCl 3) 0. 15 moles HCl ICS 11

Answer 3 3) 1500 m. L x 1 L = 1000 m. L 1. 5 L x 0. 10 mole HCl = 0. 15 mole HCl 1 L (Molarity factor) ICS 12

Test question 4 How many grams of KCl are present in 2. 5 L of 0. 50 M KCl? 1) 1. 3 g 2) 5. 0 g 3) 93 g ICS 13

Answer 4 3) 2. 5 L x 0. 50 mole x 74. 6 g KCl = 93 g KCl 1 L 1 mole KCl ICS 14

Test question 5 How many milliliters of stomach acid, which is 0. 10 M HCl, contain 0. 15 mole HCl? 1) 150 m. L 2) 1500 m. L 3) 5000 m. L ICS 15

Answer 5 2) 0. 15 mole HCl x 1 L soln x 1000 m. L 0. 10 mole HCl 1 L (Molarity inverted) = 1500 m. L HCl ICS 16

Test question 6 How many grams of Na. OH are required to prepare 400. m. L of 3. 0 M Na. OH solution? 1) 12 g 2) 48 g 3) 300 g ICS 17

Answer 6 2) 400. m. L x 1 L = 0. 400 L 1000 m. L 0. 400 L x 3. 0 mole Na. OH x 40. 0 g Na. OH 1 L 1 mole Na. OH (molar mass) = 48 g Na. OH ICS 18