The summary of the normal and molar solution preparation in pharmaceutical chemistry

1. Experiment No:02-Preparation of normal & molar
solution
Molarity & Normality
Solute is a chemical substance which is dissolved in a solution.
It can be a solid or liquid.
There are two common types of “standard solutions” in
chemistry:
• Molar solutions
• Normal solutions
Both of these solutions are concentrations (or strengths) of a
particular component (solute) that is dissolved in a solvent.
A molar solution is an aqueous solution that contains 1 mole
(gram-molecular weight) of solute in 1 liter of the solution.
A normal solution contains one equivalent (gram-equivalent
weight) of solute per liter of solution.

2. Percent solution
For acid-base reactions, an equivalent is the amount
of a reactant that produce or consume one mole of
hydrogen ions (using the Bronsted-Lowry definition).
So, for example, a mole of HCl or NaOH is one
equivalent, but a mole of H2SO4 or Ca(OH)2 is two
equivalent. Thus, a six molar (6M) sulfuric acid
solution is twelve-normal (12N).
Percent solution is the solution expressed in the unit
%. It may be (a) percentage of weight by volume-w/v,
(b) percentage of volume by volume-v/v, and (c)
percentage of weight by weight-w/w .

3. Preparation of 50ml 0.2M NaHCO3 solution
• Calculation: Here, M= 84gm
S= 0.2M, V= 50/1000 L= 0.05L
W=?
We know, W= SMV= 0.2840.05= 0.84gm
So, 0.84gm of NaHCO3 is dissolved in 50ml
distilled water to prepare 0.2M solution.

4. Preparation of 0.2M and 0.2N 100ml oxalic acid solution
Calculation:
To prepare 0.2M 100ml oxalic acid-
W=
=2.52gm
To prepare 0.2N 100ml oxalic acid
W=
= =1.26gm
So, 2.52 gm oxalic acid was dissolve in 100ml
solution & 1.26gm of oxalic acid was dissolved in
100ml solution to prepare 0.2M & 0.2N solution.

5. Preparation of 2% w/v 50 ml NaOH solution & express it in molarity & normality
1gm of NaOH is dissolved in 50ml distilled water
to prepare 2% w/v solution.
We know, S===0.5M
Again,
S== =0.5 N

6. Preparation of HCl solutions of various strengths
1. What is molarity of 37% HCl solution?
In 37% conc HCl,
mass of HCl in 1000ml=1.19g/ml x1000ml x37%=440.3g.
Molarity=440.3g/36.46g =12.07 M

7. Preparation of HCl solutions of various strengths
2. How do you get 10% w/v HCl from 37 % HCl?
What is the Molarity of this solution?
mass of HCl in 100ml=44.03g.
Volume required for 10 gm HCl-
(100/44.03 x 10)ml = 22.7ml of conc HCl.
Add water to make up to 100ml to get 10%w/v HCl
Molarity calculation:
22.7 ml contains (44.03/100)*22.7= 9.99gm
S= = (9.99*1000)/(100*36.46)= 2.74 M

8. Preparation of HCl solutions of various strengths
3.How do you get 10% v/v HCl from 37 % HCl? What is the molarity of
this solution?
(100*10/37) or 27.03 ml HCl will be taken from the bottle containing 37% HCl
and water will be added upto 100 ml.
10ml contains 11.9 g
11.9 g in 100 ml, so, 1000 ml contains 119 g
Molarity= 119/36.46= 3.26 M
4. If 10ml HCl is taken from 37 % HCl & dissolved upto 100ml distilled
water, then what is the molarity of this solution?
[V1S1 = V2S2]
M=(10*12.07)/100= 1.2 M
Task 1: What is the Molarity of 98% H2SO4?
Sulfuric acid (98%),Density: 1.83 g/cm³
Task 2: Prepare 100ml of 1N & 1M sodium carbonate solution.