The document provides information about various concentration units used in chemistry such as molarity, molality, normality, formality, percentage solutions, parts per million (ppm), specific gravity, and dilution calculations. It defines each term, provides examples of calculations using the relevant formulas, and explains how to perform dilutions to make solutions with lower concentrations from more concentrated stock solutions. Key formulas covered include those for molarity (M= moles solute/L solution), molality (m= moles solute/kg solvent), and dilution (C1V1=C2V2).
3. Mole
In how many grams ther would be 6.022 X 10 23
Moles = weight of solute (gm)
M.W. (gm/mol)
4. Molarity
Concept: Moles of solute in 1 liter of solution
M= Moles of solute (mol)
Volume of solution (L)
M= Mass of solute(gm) X 1000
M.W. of solute (gm/mol) X Volume of solution (ml)
Concept: Moles of solute in 1 liter of solution
M= Moles of solute (mol)
Volume of solution (L)
M= Mass of solute(gm) X 1000
M.W. of solute (gm/mol) X Volume of solution (ml)
5. Find the molarity of solution prepared by dissolving 6.75 g of NaCl into 452 ml
of D.W. [ M.W. of NaCl = 58.4 g/mol ]
M= Mass of solute(gm) X 1000
M.W. of solute (gm/mol) X Volume of solution (ml)
M= 6.75 X 1000
58.4 X 452
=0.2557 M
6. How many grams of MgCl2 is required to make 500ml 0.5 M MgCl2? [ M.W. of
MgCl2 =95.21 gm/mol]
M= Mass of solute(gm) X 1000
M.W. of solute (gm/mol) X Volume of solution (ml)
0.5 = x X 1000
95.21 X 500
X = 0.5 X95.21 X 500
1000
=23.802 gm
7. What is molarity of 250 ml solution containing 0.35 moles of NaCl? [ M.W. of
NaCl = 58.4 g/mol ]
M= Moles of solute (mol)
Volume of solution (L)
= 0.35
250/1000
= 0.35
0.25
= 1.4 M
8. Molality
Concept: moles of solute in 1 kg of solvent
M= Moles of solute (mol)
Mass of solvent (kg)
M= Mass of solute (gm) X 1000
M.W. of solute (gm/mol) X mass of solvent (gm)
9. What is molality when 20 gm of NaOH is dissolved in 500 gm of water?
M= Mass of solute (gm) X 1000
M.W. of solute (gm/mol) X mass of solvent (gm)
= 20 X 1000
40 X 500
= 1 N
10. Normality
Gram equivalent of solute in 1 liter of solution
N= Gram eq. of solute ( mol eq.)
Volume of solution (L)
N= Mass eq. of solute (gm) X 1000
Eq. mass of solute (gm/mol) X volume of solution (ml)
N= M X acidity/basicity
N= M X valancy
11. Calculate the normality of 1.80 g H2C2O4 dissolved in 150 ml of solution.
[M.W. =90 gm/mol]
0.267 N
12. Formality
Concept is similar to molar but it refers to original chemical formulation.
F = mass of solute (gm) X 1000
Formula M.W. weight (gm/mol) X volume of solution (ml)
13. %W/W (a.k.a. % by weight)
Concept: Mass of solute present in 100 gm of solution
%w/w= Mass of solute (gm) X 100
Mass of solution(gm)
14. %V/V (a.k.a. % by strength)
Concept: Volume of solute present in 100 ml of solution
% v/v = volume of solute (ml) X 100
Volume of solution (ml)
15. %w/v (a.k.a. % by volume)
Concept: Mass of solute present in 100 ml of solution
%w/v = Mass of solute (gm) X 100
Volume of solution (ml)
16. PPM ( parts per million)
Concept: Mass of solute per million parts of mass of solution
ppm= mass of solute(mg)
Volume of solution (L)
ppm= mass of solute (ug)
Volume of solution (ml)
18. Dilution
Dilution is done when we want to prepare solution with lower concentration
from the solution with higher concentration
The formula for dilution:C1V1=C2V2
Where, C1= concentration of stock
V1= volume of stock
C2= concentration of working solution
V2= volume of working solution
Note: concentration of both solution should be same.
19. How many ml of 2.5M NaOH is required to make 525ml of 0.15 M NaOH?
M1V1=M2V2
2.5 X x= 0.15 X 525
X= 0.15 X 525
2.5
= 31.5 ml
31.5 ml stock+493.5 ml D.W =525 final solution
20. How much of stock of 20 X SSC buffer we need to make 200 ml 2X SSC buffer.
C1V1=C2V2
20 X x = 2 X 200
x= 2 X 200
20
= 20 ml
20 ml stock
+ 180 ml D.W.
200 ml working