read this Conc by molarity measurment hindi

1. 11
:
Mr. praveen vyas Regents Chemistry
http://www.hindinext.com

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Molarity: the most common solution concentration
measurement is defined as follows:
M =M = molesmoles
Which means Molarity = # of moles per liter of solution
Capital M

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MolarityMolarity
Let’s Consider the following Problem:
What is the molarity of a solution in which 50g of
CuSO4 is dissolved in water to make a 2 liter solution.
Step 1. What’s the formula for MOLARITY?
M = moles of solute
liters of solution

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MolarityMolarity
Step 2. We need to determine how many moles we have
in 50g of CuSO4 .
How do we do that? Remember the mole Hole?
- 1. Determine the molar mass of CuSO4
(159.5g/mol)
- 2. Calculate the # of moles of CuSO4
Since we’re going into the mole hole we needSince we’re going into the mole hole we need
to DIVIDE # of grams by molar massto DIVIDE # of grams by molar mass
(50/159.5) = 0.31 moles

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MolarityMolarity
Step 3. Calculate the Molarity of the solution using
the formula for MOLARITY
M = moles of Solute
liters of solution
How many moles do we have?
How many liters do we have?
So M = 0.31 moles
2 liters M = 0.155 mol/liter
0.31 mol
2 L

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Making a 1M NaCl SolutionMaking a 1M NaCl Solution
1. Weigh out and add 1 mole of NaCl to a volumetric Flask
2. Add a small amount of solvent to dissolve the salt
3. Once the solute3. Once the solute
is dissolved addis dissolved add
additionaladditional solventsolvent
up to theup to the 1 liter1 liter
markmark

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ProblemProblem
What is the molarity of a solution made by dissolving 2.1
liters of NH3 (g) in sufficient water to make 2 liters of
solution?
1.Calculate how many moles of ammonia there are
# of moles = volume = 2.1 liters = 0.094 mol
molar vol 22.4 l/mol
What do we need to begin calculating the Molarity of the solution?

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ProblemProblem
Step 2. Calculate the molarity of the solutionStep 2. Calculate the molarity of the solution
using the molarity equationusing the molarity equation
M = # moles = 0.094 mol = 0.047M
# liters of sol’n 2 liters
We now know the number of moles of NHWe now know the number of moles of NH33 we have (0.094) so …we have (0.094) so …

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conc dil
Molarity by dilutionMolarity by dilution
Now what if we had a solution already prepared but
Wanted to decrease the molarity to some lower precise
Value.
How could we use an already prepared higher molarity
solution to accomplish this?
Note that it is only possible to go from a more
conc. (higher molarity) solution to a less conc.
(lower molarity) solution.

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Molarity by dilutionMolarity by dilution
Recall that Molarity (M) = # moles of solute
# liters of solution
So we need to dilute the solution by adding solvent
But how much more solvent do we need to add?
To make something more dilute we’re going to need to
add more solvent to the solution.

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Molarity by dilutionMolarity by dilution
When we add more solvent to a solution do we
change the number of moles of solute in the
solution?
NO! But what does change?
We change the # of moles
per unit volume of Solvent because
the same # of moles are present but
now there’s more liquid

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Molarity by DilutionMolarity by Dilution
Let’s call the Molarity of the original solution M1 and the
New Molarity M2
The number of liters of the original solution and the new
Solution will then be designated by V1 and V2
So the molarity of the old solution is…
M1 = #moles of solute = #moles
#liters of solution V1
So… #moles = M1 x V1
Let’s see how this works …

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Molarity by DilutionMolarity by Dilution
The Molarity of the new solution will then be …
M2 = #moles of solute
#liters of solution
# moles = M2 x V2
Remember that the # of moles does not change when
We dilute solutions.

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Molarity by dilutionMolarity by dilution
So…
M1 x V1 = M2 x V2
If we know three values we can calculate the 4th
.
Let’s see how we use this relationship

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Dilution ProblemDilution Problem
How much of a 3M solution of NiCL2 is needed to
Prepare 100 ml of a 1.75M solution of NiCl2?
OK how do we solve this? What’s the equation
we’re going to use?
Right! M1V1 = M2V2
M1 = 3M; M2 = 1.75M; V2 = 0.1L (100ml)
So … 3 x V1 = 1.75 x 0.1
V1 = 0.058L or 58ml

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Molarity by dilutionMolarity by dilution
How do we make 5L of
A 1.5M sol’n of KCL
From 12.0M stock sol’n?
M1V1 = M2V2
12 x V1 = 1.5 x 5
12V1= 7.5
V1 = 7.5/12 = 0.625L
Add 4.375L H20 to a
0.625L of a 12M sol’n