Normality, Molarity, Molality
source : sagarvisma-wordpress.com
METHODS OF EXPRESSING CONCENTRATION OF SOLUTION –
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