Molarity

Introduction:- 

There are different concentration terms which are used to measure "the amount of solute in respect to solution/solvent".

Molarity is one of the popular concentration term which is used commonly in every field. 

But in other hand the term molarity never used in such experiments which is related to temperature.

Definition of Molarity

Factors affecting Molarity

1. Amount of solute present in soution:- Molarity is directly proportional to the amount of solute.if amount of solute is increases then value of Molarity also increases. 
2. Dilution of solution: Dilution means adding of solvent so that volume increases . We know very well that Molarity is inversely proportional to volume.
3. Temperature:  Molarity depends on the volume of solution( Molarity = 1/ Volume ), which is affected by temperature.
   
   
   The relationship between volume and temperature
   Volume  ∝  Temperature
   
   
   So, Relationship between molarity and temperature is,
   
   
    Molarity  ∝  1/Temperature
   
   
   *If temperature increases(↑) value of molarity decreases(↓)
   
   *If temperature decreases(↓) value of molarity increases(↑) 

Also read:-

Relation between Molarity with other concentration terms

1. Relation between Molarity and %by weight and density.
2. Relation between Molarity and Mole Fraction.
3. Relation between Molarity and Normality.
4. Relation between Molarity and molality

There are different formulas related to molarity which is used according to the problem.These formule can be understood by solving some numericals.

Example question 1:-

Use of formula:- 

 Molarity = moles of solute/ volume of solution in litre.

Calculate Molarity of a NaOH solution in which 3 mol of NaOH is dissolved in 0.24 L of solution?(class 11th level question)

Ans:- Given:- amount of solute(NaOH) =3 mol

                        Volume of solution = 0.24 L

           To find:- Molarity

Molarity = 3 /0.24 = 12.5 M

Example question 2:-

Use of formula:- 

 Molarity = weight of solute × 1000/ molar mass of solute × volume of solution (ml)

Calculate Molarity of a NaoH solution in which 3 gram of NaOH is present in 400 ml of solution?(Class 11th level question)

Ans:- Given:- weight of solute(NaOH) =3 gm

          Volume of solution= 400 ml

     Molarity = moles of solute/ volume of solution (L)

Since, moles = given weight/ molar weight

Molar weight of NaOH= Na +O+ H = 23+16+1= 40

So, Molarity = 3 × 1000/ 40 × 400

                      = 3/16 

                      =0.1875 M

Example question 3:-

Use of formula:-

M1V1=M2V2

Ques 1. 200 ml of 0.1 molar  CH₃COOH solution is converted to  800 ml by adding water . What is the new molarity. (12th Board level question)

Sol:- we know that, M=mole/volume 

Or MV=mole

Key point:- Solution is diluted with adding solvent hence, mole of solute present in solution is constant.

                

Moles before dilution=MV1

Moles after dilution= M2V2

 

Moles do not change = M1V1=M2V2

0.1×200=M2.800

20/800 = 1/40 =0.025 M

Hence the new molarity of solution is 0.025 M

Ques 2:-200 ml of 0.1 M  CH₃COOH solution is mixed with  400 ml of 0.5 M CH₃COOH  solution, what will be the new molarity of resultant solution . (NEET level question)

Solution:- Here, V1=200 ml

                             M1= 0.1

                             V2 = 400 ml

                             M2= 0.5

When solution 1 is mixed with solution 2

  

No of moles of solution 1 = M1V1

                                              = 0.1×200

                                              = 20

Since M1= Moles/V1

By rearrangement 

Moles= M1V1

Similarly Moles of solution 2 = M2V2

                                                    = 0.5×400

                                                    = 200

Total volume of resultant solution = V1+V2

                                                             =200+400

                                                             = 600

Molarity of resultant solution  = moles of solution 1+ moles of solution 2/ Total volume of resultant solution

Molarity of resultant solution= M1V1+M2V2/V1+V2

=20+200/600

=220/600

=0.37 mol L⁻¹