Molar and Normal Solutions

 

Molar and Normal Solutions

¨  During chemical analysis different types of solutions at different concentration are used.

¨  Commonly used solutions are Molar solutions and Normal Solutions

Molar Solutions

¨  For the preparation of molar solutions we dissolve gram molecular weight of the solute per  liter of solution.

¨  To prepare 1 molar, 1 liter solution we need to dissolve the solute equal to the molecular weight of the solute in grams.

¨  Formula

¨  Molarity= No. of moles/per liter of solution

¨  or

¨  Mass in gram= Molarity x molecular weight x required volume

                                                             1000

Examples

Example 1

¨  Preparation of 1M solution of H2SO4

¨  Molecular Weight of H2SO4 = 2+32+64= 98

¨  Mass in gram= Molarity x molecular weight x required volume

                                                             1000

¨  Mass in gram= 1 x 98.08 x 1000  = 98.08 gram

                                        1000

Now convert mass into volume by using following formula

¨  Density = Mass   

                    Volume

Re arrange the formula, Volume = Mass

                                                            density

Density of H2SO4 = 1.83

Volume = 98.08/ 1.83 = 53.6 ml

¨  The above said volume is for 100% pure H2SO4, however the available H2SO4 are 98% pure, hence the volume required will become

¨   53.6 x 100  = 54.7 ml

       98

¨  To prepare 1 liter, 1 molar solution of H2SO4 we will take 54.7 ml.

Example 2

¨  1M solution of NaOH

¨  Molecular weight of NaOH = 23+16+1= 40

¨  We need 40 gram of NaOH to make 1 liter solution of 1M concentration of NaOH

¨  Mass in gram= Molarity x molecular weight x required volume

                                                             1000

Mass in gram = 1 x 40 x 1000     = 40

                                    1000

Normal Solutions

¨  To prepare normal solution we dissolve gram equivalent weight of the solute per liter of solution.

¨  It means to prepare 1 liter solution of 1 normal, we need to dissolve the solute equal to the gram equivalent weight of the solute.

¨  To find equivalent weight of the solute we need to divide molecular weight by its valency.

equivalent weight

¨  Normality = gram equivalent weight

                           per liter of solution

¨  or

¨  Mass in gram=

¨  Normality x gram equivalent weight x required volume

                                                             1000

¨  Gram Equivalent weight= Molecular weight

                                                          valency

Examples

     Example 1

¨  1N solution of H2SO4

¨  Molecular weight of H2SO4= 2+32+64= 98.08

¨  Valency of H2SO4 = 2 (2 hydrogen Ion)

¨  Equivalent weight of H2SO4 =98.08/2 = 49.04

¨  Mass in gram=

¨  Normality x gram equivalent weight x required volume

                                                             1000

¨  Mass in gram= 1 x 49.04 x 1000  = 49.04 gm

                                     1000

¨  Now convert mass into volume by using following formula

¨  Density = Mass  

                    Volume

Re arrange the formula, Volume = Mass

                                                            density

Density of H2SO4 = 1.83

Volume = 49.04/ 1.83 = 26.8 ml

¨  The above said volume is for 100% pure H2SO4, however the available H2SO4 are 98% pure, hence the volume required will become

¨   26.8 x 100  = 27.4 ml

       98

¨  To prepare 1 liter, 1 normal solution of H2SO4 we will take 27.4 ml.

Example 2

¨  1N solution of NaOH

¨  To prepare 1N solution of NaOH proceed as that of 1M NaOH solution because the gram equivalent weight of NaOH is equal to gram molecular weight of NaOH

Presented by

Muhammad Atif

PhD Scholar Pharmaceutical Sciences

Abdul Wali Khan University Mardan