Introduction :
The amount of substance present in unit amount of the solution is called concentration of the solution. Generally, concentration of a solution is expressed in
(a) molarity, M (b) molality, m or (c) normality, N
Molarity:
“The number of gram molecular weight of the solute present in 1000 cm3 (or 1 dm3) of the solution is called molarity”. It is denoted by the symbol M.
Suppose 1 mole of Oxalic acid crystals (H2C2O4. 2H2O, Molecular weight 126g) is dissolved in 1dm3 of the solution. Molarity of the solution is 1.
Molarity of the solutions can be calculated from the expression
Formula for Molarity = (mass/dm3) / molecular weight
Case 2:
Suppose ‘x’ molar solution of Oxalic acid is asked to prepare. Then, the weight of the Oxalic acid corresponding to ‘x’ mole of Oxalic acid is used in making x molar solution.
That is weight of Oxalic acid required is = ‘x’ moles X molecular weight of Oxalic acid
= molarity X 126g
This much of Oxalic acid should be dissolved in 1 dm3 of solvent.
Case 3:
For the making of ‘x’ molar solutions of Oxalic acid of a volume, say 100ml, the weight of the oxalic acid required is found using the below relation.
Case 4:
If a liquid reagent of certain % assay and density is given in the making of certain molar solutions, then the strength of the given liquid reagent should be determined first. This is done by using the relation.
Strength = number of moles / dm3
= (weight in g / Molecular weight) / dm3
= density / Molecular weight
Now, the strength of the liquid reagent given is used in the making of the solution of required strength and volume using the below relation.
M1.V1 = M2 V2
Where:
M1, M2 are the molarities of the given liquid reagent and solution to be prepared
V1, V2 are the volumes of the given liquid reagent and solution to be prepared, respectively.
Since, Equivalent weight and Molecular weight are related as:
So, there exists a relation between the molarity and normality as below:
Molarity = (mass/dm3) / Mol. Wt
= (mass/dm3) / [Eq. wt x valence (number of electron exchange)]
Molarity = Normality / valence or
Normality = Molarity X valence is used in the making of N solutions from M solutions.
The amount of substance present in unit amount of the solution is called concentration of the solution. Generally, concentration of a solution is expressed in
(a) molarity, M (b) molality, m or (c) normality, N
Molarity:
“The number of gram molecular weight of the solute present in 1000 cm3 (or 1 dm3) of the solution is called molarity”. It is denoted by the symbol M.
Making molar solutions Case1 and 2
Case 1:Suppose 1 mole of Oxalic acid crystals (H2C2O4. 2H2O, Molecular weight 126g) is dissolved in 1dm3 of the solution. Molarity of the solution is 1.
Molarity of the solutions can be calculated from the expression
Formula for Molarity = (mass/dm3) / molecular weight
Case 2:
Suppose ‘x’ molar solution of Oxalic acid is asked to prepare. Then, the weight of the Oxalic acid corresponding to ‘x’ mole of Oxalic acid is used in making x molar solution.
That is weight of Oxalic acid required is = ‘x’ moles X molecular weight of Oxalic acid
= molarity X 126g
This much of Oxalic acid should be dissolved in 1 dm3 of solvent.
Making molar solutions Case3 ,4 and relation between molarity & normality
Case 3:
For the making of ‘x’ molar solutions of Oxalic acid of a volume, say 100ml, the weight of the oxalic acid required is found using the below relation.
Case 4:
If a liquid reagent of certain % assay and density is given in the making of certain molar solutions, then the strength of the given liquid reagent should be determined first. This is done by using the relation.
Strength = number of moles / dm3
= (weight in g / Molecular weight) / dm3
= density / Molecular weight
Now, the strength of the liquid reagent given is used in the making of the solution of required strength and volume using the below relation.
M1.V1 = M2 V2
Where:
M1, M2 are the molarities of the given liquid reagent and solution to be prepared
V1, V2 are the volumes of the given liquid reagent and solution to be prepared, respectively.
Since, Equivalent weight and Molecular weight are related as:
So, there exists a relation between the molarity and normality as below:
Molarity = (mass/dm3) / Mol. Wt
= (mass/dm3) / [Eq. wt x valence (number of electron exchange)]
Molarity = Normality / valence or
Normality = Molarity X valence is used in the making of N solutions from M solutions.
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