Normality, Molarity, Molality source : sagarvisma-wordpress.com

## 1) Concentration  in gram/litre  :

“It is equal to the number of grams  of the solute  present in one liter of solution.”

S (Conc.  in gram/litre) = mass of solute (gram)/Volume of solution (liter)

S =w/V(l)

## 2) Mass Percentage :

“It is equal to the mass of the solute in grams present in 100 grams  of solution.”

mass % = (mass of solute x 100)/mass of solution

## 3)  Volume   Percentage :

“It is equal to the volume  of the one component present in 100 parts of the solution by volume.”

volume % = (volume of solute x 100)/volume of solution

## 4) Normality :

“Number  of gram  equivalents  of solute  present in one liter of solution is called normality of solution”.

Normality (N)= Number  of gm. equivalents of solute/volume of solution( liter)

Number  of gm.  equivalents of solute = w/E =weight of solute/Equivalent weight of solute

N = w / EV(l)

N/10 ( decinormal )= When 1/10 gm.equivalents of solute  are present in one liter of solution,then solution is  decinormal.

N/100 ( centi normal )= When 1/100 gm.equivalents of solute  are  present in one liter of solution,then solution is centinormal.

N/1000 ( milli normal )= When 1/10oo gm.equivalents of solute are  present in one liter of solution,then solution is  milli normal.

## 5)  Molarity :

“Number  of moles of solute present in one liter  of solution is called molarity of solution”.

Molarity  (M)= Number  of moles of solute/volume of solution(liter)

Number  of moles  of solute = w/m =weight of solute/molecular weight of solute

M =w/m V(l)

M/10 ( decimolar ) = When 1/10 moles of solute are  present in one liter of solution,then solution is decimolar.

M/100 ( centi molar )= When 1/100 moles  of solute are  present in one liter of solution,then solution is centimolar.

M/1000 ( milli molar )= When 1/10oo moles of solute are  present in one liter of solution,then solution is millimolar.

## Relation between  Molarity & Normality :

Normality/ Molarity = molecular weight /Equivalent weight

Q.  6 gm. of a solute is present in 500 ml of solution. what is the concentration  of solution in gm/liter ?
Solution –

w=6 gm. : V= 500 ml. =0.5 liter

S = w/V (l)

=6/0.5

S =12 gm/liter
Q. Calculate the normality of the solution containing  5 gram NaOH dissolved in 250 ml. aqueous solution.

Solution :

w =5 gm. : V =250 ml. =0.250 liter

E of  NaOH =23+16 +1 =40

N =w/ EV(l)

= 5/ (40 x 0.250) = 0.5 gm. equi. /liter