Advanced Chemistry Introduction to Molarity Ms. Grobsky

Advanced ChemistryIntroduction to Molarity

Ms. Grobsky

Real-Life Applications of Concentration and Molarity

• Prescription drugs in the correct concentration make you better– In higher concentration, they can kill you

• Pesticides must be in proper concentrations• Food additives must be in correct

concentrations• A driver is legally impaired at 0.08 mg/mL

blood alcohol content

Chemistry and Molarity

• For these reasons, chemists need to make solutions that have precise concentrations

• But, how do they do that?– Chemists control

the concentration of chemicals using the concept of molarity

Molarity and Solutions

• Scientists use molarity to describe concentrations in chemistry

• The concentration of a solution tells you how much solute that is dissolved in a given amount of solution (solute + solvent).

• The molarity is the concentration of a solution

Molarity = Moles of solute

Liter of solution

Molarity

• Conversion factors can be used to convert between amounts of solute and the solvent

• Molarity does NOT mean number of moles!

Sample ProblemSample Problem

• 4.0 L of solution is prepared from 2.255 moles of NaCl. Calculate the molarity.

ALWAYS START WITH WHAT YOU KNOW!

Sample ProblemSample Problem

4.0 L of solution is prepared from 2.255 moles of NaCl. Calculate the molarity.

Molarity = 2.255 moles

4.0 L

= 0.56 moles/liter

= 0.56 M (Molar)

where M means moles per 1 liter

Practice problemsPractice problems

• 250. mL of solution is prepared from 5.00 g KOH. Calculate the molarity.



• 250. mL of solution is prepared from 5.00 g KOH. Calculate the molarity.

Molarity = 5.00 g x 1 mole

56.11 g

---------------------------

0.250 L

= 0.356 M


• How many moles are there in 205. mL of a 0.172 M solution?– Remember, 205 mL of 0.172 M solution

does NOT mean there are .172 moles!



• How many moles are there in 205. mL of a 0.172 M solution?

0.205 L x 0.172 moles = 0.0353 moles 1 L


• How many grams NaCl are there in 250.0 mL of 0.500 M solution?



• How many grams NaCl are there in 250.0 mL of 0.500 M solution?

0.250 L x 0.500 moles x 58.5 g = 7.31 g

1 L 1 mole


• How many grams of NaCl must be used to prepare a 100.0 mL of 0.250 M solution?



• How many grams of NaCl must be used to prepare a 100.0 mL of 0.250 M solution?

0.1000 L x 0.250 moles x 58.5 g = 1.46 g

1 L 1 mole

Preparation of SolutionsPreparation of Solutions

• I could ask you to make me a 2.25 M solution of HCl using 1.45 liters of water, and you could do it. How?

• Give directions for the preparation of 2.50 L of a 1.34 M NaCl solution

Practice problemPractice problem

• Describe how you would prepare 100.0 mL of a 0.200 M solution of CoCl2.– Find moles of CoCl2– Convert to grams

– Mass amount of grams of CoCl2 on a balance

– Dissolve in water

– Transfer to a 100 mL volumetric flask and fill to the line

Making Molar

Solutions From Liquids(More accurately, from

stock solutions)

Making Molar Solutions from LiquidsMaking Molar Solutions from Liquids

• Not all compounds are in a solid form

• Acids are purchased as liquids (“stock solutions”)

– Yet, we still need a way to make molar solutions of these compounds

• The procedure is similar to before

– Use pipette to measure moles (via volume)

– Use volumetric flask to measure volume

• Now we use the equation M1V1 = M2V2

– 1 is starting (concentrated conditions)

– 2 is ending (dilute conditions)

The Dilution FormulaThe Dilution FormulaExample: If we have 1 L of 3 M HCl, what is

M if we dilute acid to 6 L?

M1 = 3 mol/L, V1 = 1 L, V2 = 6 L

M1V1 = M2V2, M1V1/V2 = M2

M2 = (3 mol/L x 1 L) / (6 L) = 0.5 MWhy does the formula work? Because we are equating mol to mol:

M1V1 = 3 mol

V1 = 1 LM1 = 3 M

V2 = 6 LM2 = 0.5 M

M2V2 = 3 mol

Practice problemsPractice problemsProblem 1What volume of 0.5 M HCl can be prepared

from 1 L of 12 M HCl?

M1 = 12 mol/L, V1 = 1 L, M2 = 0.5 L

M1V1 = M2V2

M1V1/M2 = V2

V2 = (12 mol/L x 1 L) / (0.5 L) = 24 L


Problem 21 L of a 3 M HCl solution is added to 0.5 L of a 2 M HCl

solution. What is the final concentration of HCl?

(Hint: first calculate total number of moles and total number of L)

# mol = (3 mol/L)(1 L) + (2 mol/L)(0.5 L)= 3 mol + 1 mol = 4 mol

# L = 1 L + 0.5 L = 1.5 L# mol/L = 4 mol / 1.5 L = 2.67 mol/L


• How many mL of a 14 M stock solution must be used to make 250 mL of a 1.75 M solution?

• You have 200 mL of 6.0 M HF. What concentration results if this is diluted to a total volume of 1 L?

• 100 mL of 6.0 M CuSO4 must be diluted to what final volume so that the resulting solution is 1.5 M?


• What concentration results from mixing 400 mL of 2.0 M HCl with 600 mL of 3.0 M HCl?

• What is the concentration of NaCl when 3 L of 0.5 M NaCl are mixed with 2 L of 0.2 M NaCl?

• What is the concentration of NaCl when 3 L of 0.5 M NaCl are mixed with 2 L of water?


• Water is added to 4 L of 6 M antifreeze until it is 1.5 M. What is the total volume of the new solution?

• There are 3 L of 0.2 M HF. 1.7 L of this is poured out, what is the concentration of the remaining HF?

Dilution problems (1-6, 6 two ways)Dilution problems (1-6, 6 two ways)1.

M1 = 14 M, V1 = ?, M2 = 1.75 M, V2 = 250 mLV1 = M2V2 / M1 = (1.75 M)(0.250 L) / (14 M)V1 = 0.03125 L = 31.25 mL

2.M1 = 6 M, V1 = 0.2 L, M2 = ?, V2 = 1 LM2 = M1V1 / V2 = (6 M)(0.2 L) / (1 L)M2 = 1.2 M

3.M1 = 6 M, V1 = 100 mL, M2 = 1.5 M, V2 = ?V2 = M1V1 / M2 = (6 M)(0.100 L) / (1.5 M)V2 = 0.4 L or 400 mL

Dilution problems (4 - 6)Dilution problems (4 - 6)4. # mol = (2.0 mol/L)(0.4 L) + (3.0 mol/L)(0.6 L)

= 0.8 mol + 1.8 mol = 2.6 mol # L = 0.4 L + 0.6 L # mol/L = 2.6 mol / 1 L = 2.6 mol/L

5. # mol = (0.5 mol/L)(3 L) + (0.2 mol/L)(2 L) = 1.5 mol + 0.4 mol = 1.9 mol

# mol/L = 1.9 mol / 5 L = 0.38 mol/L6. # mol = (0.5 mol/L)(3 L) + (0 mol/L)(2 L)

= 1.5 mol + 0 mol = 1.5 mol # mol/L = 1.5 mol / 5 L = 0.3 mol/LOr, using M1V1 = M2V2,

M1 = 0.5 M, V1 = 3 L, M2 = ?, V2 = 5 L

7.M1 = 6 M, V1 = 4 L, M2 = 1.5 M, V2 = ?

V2 = M1V1 / M2 = (6 M)(4 L) / (1.5 M)

V2 = 16 L

8. The concentration remains 0.2 M, both volume and moles are removed when the solution is poured out. Remember M is mol/L. Just like the density of a copper penny does not change if it is cut in half, the concentration of a solution does not change if it is cut in half.

Dilution problems (7, 8)Dilution problems (7, 8)

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Advanced Chemistry Introduction to Molarity Ms. Grobsky