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Concentration of SoluteConcentration of SoluteConcentration of SoluteConcentration of Solute

The amount of solute in a The amount of solute in a solution is given by its solution is given by its concentrationconcentration.

Molarity (M) = moles soluteliters of solution

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1.0 L of 1.0 L of water was water was

used to used to make 1.0 L make 1.0 L of solution. of solution. Notice the Notice the water left water left

over.over.

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PROBLEM: Dissolve 5.00 g of PROBLEM: Dissolve 5.00 g of NiClNiCl22•6 H•6 H22O in enough water to O in enough water to make 250 mL of solution. make 250 mL of solution. Calculate the Molarity.Calculate the Molarity.

PROBLEM: Dissolve 5.00 g of PROBLEM: Dissolve 5.00 g of NiClNiCl22•6 H•6 H22O in enough water to O in enough water to make 250 mL of solution. make 250 mL of solution. Calculate the Molarity.Calculate the Molarity.

Step 1: Step 1: Calculate moles Calculate moles of NiClof NiCl22•6H•6H22OO

5.00 g • 1 mol

237.7 g = 0.0210 mol

0.0210 mol0.250 L

= 0.0841 M

Step 2: Step 2: Calculate MolarityCalculate Molarity

[NiClNiCl22•6 H•6 H22OO ] = 0.0841 M

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Step 1: Change mL to L.Step 1: Change mL to L.

250 mL * 1L/1000mL = 0.250 L250 mL * 1L/1000mL = 0.250 L

Step 2: Calculate.Step 2: Calculate.

Moles = (0.0500 mol/L) (0.250 L) = 0.0125 molesMoles = (0.0500 mol/L) (0.250 L) = 0.0125 moles

Step 3: Convert moles to grams.Step 3: Convert moles to grams.

(0.0125 mol)(90.00 g/mol) = (0.0125 mol)(90.00 g/mol) = 1.13 g1.13 g

USING MOLARITYUSING MOLARITYUSING MOLARITYUSING MOLARITY

moles = M•Vmoles = M•V

What mass of oxalic acid, What mass of oxalic acid, HH22CC22OO44, is, is

required to make 250. mL of a 0.0500 Mrequired to make 250. mL of a 0.0500 Msolution?solution?

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Learning Check

How many grams of NaOH are required to prepare 400. mL of 3.0 M NaOH solution?

1) 12 g

2) 37 g

3) 300 g

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An An IDEAL SOLUTIONIDEAL SOLUTION is one is one where the properties depend where the properties depend only on the concentration of only on the concentration of solute.solute.

Need conc. units to tell us the number Need conc. units to tell us the number of solute particles per solvent of solute particles per solvent particle.particle.

The unit “molarity” does not do this!The unit “molarity” does not do this!

Concentration UnitsConcentration Units

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Two Other Concentration Two Other Concentration UnitsUnits

grams solutegrams solutegrams solutiongrams solution

MOLALITY, mMOLALITY, m

% by mass% by mass = =

% by mass% by mass

m of solution = mol solute

kilograms solvent

8Calculating Calculating ConcentrationsConcentrations

Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g of Hof H22O. Calculate molality and % by mass of O. Calculate molality and % by mass of

ethylene glycol.ethylene glycol.

9Calculating Calculating ConcentrationsConcentrations

Calculate molalityCalculate molality

Calculate molalityCalculate molality

Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g

of Hof H22O. Calculate m & % of ethylene glycol (by mass).O. Calculate m & % of ethylene glycol (by mass).

Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g

of Hof H22O. Calculate m & % of ethylene glycol (by mass).O. Calculate m & % of ethylene glycol (by mass).

conc (molality) = 1.00 mol glycol0.250 kg H2O

4.00 molalconc (molality) = 1.00 mol glycol0.250 kg H2O

4.00 molal

%glycol = 62.1 g

62.1 g + 250. g x 100% = 19.9%%glycol =

62.1 g62.1 g + 250. g

x 100% = 19.9%

Calculate weight %Calculate weight %

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Learning Check

A solution contains 15 g Na2CO3 and 235 g of

H2O? What is the mass % of the solution?

1) 15% Na2CO3

2) 6.4% Na2CO3

3) 6.0% Na2CO3

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Using mass %

How many grams of NaCl are needed to prepare 250 g of a 10.0% (by mass) NaCl solution?

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Try this molality problem

• 25.0 g of NaCl is dissolved in 5000. mL of water. Find the molality (m) of the resulting solution.

m = mol solute / kg solvent

25 g NaCl 1 mol NaCl

58.5 g NaCl= 0.427 mol NaCl

Since the density of water is 1 g/mL, 5000 mL = 5000 g, which is 5 kg

0.427 mol NaCl

5 kg water= 0.0854 m salt water