Normality, Molarity,Molality

Normality, Molarity,Molality

source :khan academy

Molality  :

“Number of  moles of solute present in one kg. of solvent  is called molality “.

Molality  = Number  of moles of solute/weight  of solvent(Kg.)

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

    =w/m W(Kg.)

Deci molal  = When 1/10 moles  of solute are present in one kg. of solvent ,then solution is deci molal.

Centi molal  = When 1/100 moles  of solute are  present in one Kg. of solvent ,then solution is centi molal.

milli molal  = When 1/1000  moles of solute are  present in one Kg. of solvent  then  solution is milli molal.


Mole fraction:

” Number  of moles of the substance  divided by total  number of moles of Solution is called mole fraction “.

If  three components A, B & C are  present,
then Number  of moles of A =  nA
Number  of moles of B = nB
Number  of moles of C = nC
Total no. of moles =  nA+nB+nC
Mole fraction of A = nA/(nA + nB+ nC)

XA = (wA/mA) /[ (wA/mA)+(wB/mB)+(wC/mC)]

XB =(wB/mB) /[ (wA/mA)+(wB/mB)+(wC/mC)]

XC  = (wC/mC) /[ (wA/mA)+(wB/mB)+(wC/mC)]
(XA+XB+XC = 1) mole fraction is always unity.

Concentration in ppm (parts per million) –

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

concentration  in ppm =weight of solute (mg.)/ volume of solution (l)

Formality :

“Formality is equal to the number of formula masses present in one liter of the Solution.”

F = number of formula masses of solute / Volume of solution (L)

F = weight of solute in gm. /  (formula mass of solute x volume of solution in liter)

It is used to express the concentration of ionic components such as NaCl , KBr , KI etc.

Q.  7.45 gm of  KCl is dissolved in 100 gm of  H2O. calculate mole fraction of KCl ?

Solution :

m of KCl =39 +35.5 =74.5
moles of KCl =w/m= 7.45 /74.5 =0.1

moles of  H2O =w/m =100/18 = 5.55

total moles =0.1 +5.55 =5.65

mole fraction of KCl= moles  of KCl / total moles

0.1 /5.65 =0 .018

mole fraction of KCl =0.018 ans.
Q. A sugar syrup of weight 214.2 gm. contains 34.2 gm. of sugar. Calculate  (i) Molality or  Molal  Concentration. (ii) Mole fraction of sugar & water in the syrup?

molecular weight  of sugar = 342

w = 34.2 gm.

Weight  of solvent= 214.2-34.2
= 180 gm
= 0.180 kg.

Molality  = w/ m W(Kg.)

=34.2/ 342 x 0.180

   Molality  = 0.56 mole /Kg.
no. of moles of sugar (ns) =w/m 34.2/342
= 0.1
Moles of water  nw = w/m = 180/18 = 10

Total moles= 10 +0.1 =10.1

Xs = ns/total moles

=0.1/10.1= 0.0099

mole fraction of sugar  =0.0099

Xw  = nw / total moles

=10/10.1 =0.990

mole fraction of water =0.990