CBSE Class 12 Chemistry Solutions Board Exam Notes

Download CBSE Class 12 Chemistry Solutions Board Exam Notes in PDF format. All Revision notes for Class 12 Chemistry have been designed as per the latest syllabus and updated chapters given in your textbook for Chemistry in Class 12. Our teachers have designed these concept notes for the benefit of Class 12 students. You should use these chapter wise notes for revision on daily basis. These study notes can also be used for learning each chapter and its important and difficult topics or revision just before your exams to help you get better scores in upcoming examinations, You can also use Printable notes for Class 12 Chemistry for faster revision of difficult topics and get higher rank. After reading these notes also refer to MCQ questions for Class 12 Chemistry given on studiestoday

Revision Notes for Class 12 Chemistry Unit 02 Solutions

Class 12 Chemistry students should refer to the following concepts and notes for Unit 02 Solutions in Class 12. These exam notes for Class 12 Chemistry will be very useful for upcoming class tests and examinations and help you to score good marks

Unit 02 Solutions Notes Class 12 Chemistry

SOLUTION is the homogeneous mixture of two or more than two components. Most of the solutions are binary i.e. consists of two components out of which that is present in the largest quantity is called solvent & one which is present in smaller quantity called solute.
 
EXPRESSING CONCENTRATIONS OF SOLUTIONS

¶ Mass percentage: Mass of solute per 100g of solution Mass % = (mass of solute / total mass of solution) X 100
¶  Volume percentage : volume of solute per 100ml of solution Volume % = (volume of solute/ total volume of solution) X 100
 
¶  Parts per million: parts of a component per million (106) parts of the solution.
ppm = no. o f parts of the component / total no. of parts of all components of the sol. X 106
¶  Mole fraction(x): It is the ratio of no. of moles of one component to the total no. of all the components present in the solution . For binary solution :‐ the no. of moles of A and B are nA and nB respectively
so, xA= nA/nA+nB ; xB= nB/ nA+nB
In binary solution xA+ xB= 1
¶  Molarity: No. of moles of solute dissolved in one litre of solution .
Molarity(M) = moles of solute/ vol. of solution in litre
¶  Molality(m) : No. of moles of solute per kg of the solvent.
Molality(m) = moles of solute/mass of solvent in kg
Molality is independent of temp. whereas molarity is a function of temp. because vol. depends on temp. and mass does not.
 
HENRY'S LAW

It states that at a constant temp. the solubility of the gas in liquid is directly proportional to the pressure of the gas above the surface of the liquid .
It also states that the partial pressure (p) of a gas in vapour phase is proportional to the mole fraction of the gas (x) in the solution.
P=KHX
KH is Henry's law constant .
 
APPLICATION OF HENRY'S LAW

¶  To increase the solubility of CO2 in soda water and soft drinks the bottle is sealed under high pressure.
¶  To avoid bends , toxic effects of high concentration of nitrogen in the blood the tanks used Used by scuba divers are filled with air diluted with He.
 
RAOULT'S LAW :‐ it states that :
1) For a solution of volatile liquid , the partial vapour pressure of each component of the solution is directly proportional to its mole fraction present in solution.
PA= P0
A XA PB= P0B XB
The total pressure is equal to sum of partial pressure. Ptotal= PA+ PB
2) For a solution containing non‐volatile solute the vapour pressure of the solution is directly proportional to the mole fraction of the solvent.
 
 
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CBSE Class 12 Chemistry Unit 02 Solutions Notes

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Notes for Chemistry CBSE Class 12 Unit 02 Solutions

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Unit 02 Solutions Notes for Chemistry CBSE Class 12

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Unit 02 Solutions CBSE Class 12 Chemistry Notes

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Notes for CBSE Chemistry Class 12 Unit 02 Solutions

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