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# Molarity Molality Dilutions

## by Regis Komperda, Chemistry / Physics Teacher at Minooka Community High School on May 06, 2009

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## Molarity Molality DilutionsPresentation Transcript

• Concentration of Solute
• The amount of solute in a solution is given by its concentration .
Molarity ( M ) = moles solute liters of solution
• 1.0 L of water was used to make 1.0 L of solution. Notice the water left over.
• PROBLEM: Dissolve 5.00 g of NiCl 2 •6 H 2 O in enough water to make 250 mL of solution. Calculate the Molarity. Step 1: Calculate moles of NiCl 2 •6H 2 O Step 2: Calculate Molarity [ NiCl 2 •6 H 2 O ] = 0.0841 M
• Step 1: Change mL to L.
• 250 mL * 1L/1000mL = 0.250 L
• Step 2: Calculate.
• Moles = (0.0500 mol/L) (0.250 L) = 0.0125 moles
• Step 3: Convert moles to grams.
• (0.0125 mol)(90.00 g/mol) = 1.13 g
USING MOLARITY moles = M•V What mass of oxalic acid, H 2 C 2 O 4 , is required to make 250. mL of a 0.0500 M solution?
• Learning Check
• How many grams of NaOH are required to prepare 400. mL of 3.0 M NaOH solution?
• 1) 12 g
• 2) 48 g
• 3) 300 g
• The Other Concentration Unit MOLALITY, m m of solution = mol solute kilograms solvent
• Calculating Concentrations
• Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g of H 2 O. Calculate molality ethylene glycol.
• 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
• Preparing Solutions
• Weigh out a solid solute and dissolve in a given quantity of solvent.
• Dilute a concentrated solution to give one that is less concentrated.
• Dilutions
• Dilutions are especially important in the lab
• Solutions are shipped in concentrated form, but need to be diluted for use in labs
• Important:
• When you dilute a solution you are NOT changing the number of moles of solute, just volume of solvent
• The concentration of
• the sample is the SAME
• concentration as the
• original solution
• Solving a dilution problem
• You start with 1 L of a 5 M solution. How would you make this into a 2.5 M solution?
• What if you only wanted 1 L of the 2.5 M solution, and didn’t want to waste any of the 5 M solution?
• We know that any moles brought into the solution stay in the solution. So…
• How many moles do we need in our final solution?
• m = M * V
• m = 2.5 M * 1 L = 2.5 moles
• How what volume of stock solution contains that many moles?
• V = m/M
• V = 2.5 moles /5 M = .5 L
• We need to take out .5 L of stock solution then add water to make 1 L of solution
• M 1 V 1 =M 2 V 2
• This is the same process we used on the last slide, just combined into one equation.
• The first side represents the stock solution, the second side represents the diluted solution
• REMEMBER!!!! The volume you solve for is the volume of stock solution you need, this is not the overall answer to the problem!!!!
• Try these!
• If 250 mL of .10 M sodium chloride was diluted to a volume of 750 mL, what will the new concentration be?
• How would you make 50 mL of 3 M solution if you started with 30 M solution?
M 1 V 1 =M 2 V 2