Presentation
on
Normality
The strength of solution measured in
terms of gram equivalent per litre is
called normality.
It is denoted by N.
A solution having 1 gram equivalent of
the dissolved solute in 1 litre of its
solution is called normal solution.
The solution in which 1/10th gram
equivalent of solute is dissolved per litre
of its solution,is called decinormal
solution i.e ,the solution will have N/10
strength.
Normality depends on two factors:
a) Dilution
b) Temperature
Mathematically,
N = weight in grams X 1000
equivalent weight expressed in gram X V
or,
N = strength in gram/litre
gm.eq.wt of the compound
NUMERICAL
Calculate the normality of x gram of NaOH
dissolved in y ml of solution.(eq.weight of
NaOH=40)
Here,
N = weight in gram X1000
eq.wt X V
= x X 1000
40 X y
NORMALITY FORMULA
Suppose N1 and V1 are normality and
volume of one solution and N2 and V2 are
the normality and volume of another
solution,then acoording to the law of
equivalent,
N1 V1 = N2 V2
This is called normality formula.
NUMERICAL
a mL of an unknown acid required b mL
of x N NaOH to reach the equivalence
point.Then calculate the normality of
unknown acid.
Here,
According to normality formula,
N1V1=N2V2
a X x = b X N2
Normality

Normality

  • 1.
  • 2.
    The strength ofsolution measured in terms of gram equivalent per litre is called normality. It is denoted by N. A solution having 1 gram equivalent of the dissolved solute in 1 litre of its solution is called normal solution.
  • 3.
    The solution inwhich 1/10th gram equivalent of solute is dissolved per litre of its solution,is called decinormal solution i.e ,the solution will have N/10 strength. Normality depends on two factors: a) Dilution b) Temperature
  • 4.
    Mathematically, N = weightin grams X 1000 equivalent weight expressed in gram X V or, N = strength in gram/litre gm.eq.wt of the compound
  • 5.
    NUMERICAL Calculate the normalityof x gram of NaOH dissolved in y ml of solution.(eq.weight of NaOH=40) Here, N = weight in gram X1000 eq.wt X V = x X 1000 40 X y
  • 6.
    NORMALITY FORMULA Suppose N1and V1 are normality and volume of one solution and N2 and V2 are the normality and volume of another solution,then acoording to the law of equivalent, N1 V1 = N2 V2 This is called normality formula.
  • 7.
    NUMERICAL a mL ofan unknown acid required b mL of x N NaOH to reach the equivalence point.Then calculate the normality of unknown acid. Here, According to normality formula, N1V1=N2V2 a X x = b X N2