--%>

Avogadro's hypothesis Law Principle

Avogadro's hypothesis Law Principle- Berzelius, a chemist tried to correlate Dalton's atomic theory & Gay-Lussac's Law of gaseous volumes. According to his Berzelius hypothesis Equal volumes of all gases under similar conditions of temperature & pressure contain equal number of atoms e.g.

Hydrogen (1 vol) + Chlorine (1 vol)->HCl (2 vol)

Acc to Berzelius hypothesis:

Hydrogen (1/2 atom) + Chlorine (1/2 atom)->HCl (1 compound atom)

But this is indirect conflict of Dalton's atomic theory, so it was rejected.

So a new hypothesis was given by Avogardo.

According to him, An atom is a smallest particle of an element which can take part in a chemical reaction which may or may not be capable of independent existence.

molecule is the smallest particle of an element or of a compound which have an independent existence. So the smallest particle of a gas is a molecule not an atom, so the volume of gas must be related to the number of molecules rather than atoms.

According to Avogrado's Hypothesis-Equal volume of all gases under similar conditions of temperature & pressure contain equal number of molecules. This is able to explain all the gaseous reactions & now known as Avogrado's Law or Avogrado's principle.

For example-

Hydrogen (1 vol) + Chlorine (1 vol)->HCl (2 vol)

By Avogrado's hypothesis:

n molecule+n molecule gives 2n molecule

1/2molecule of both [Hydrogen + Chlorine] ->HCl (1 molecule)

 Applications of this hypothesis-

(1)In the calculation of atomicity of elementary gases-Atomicity is defined as the number of atoms of the element present in one molecule of the substance e.g. atomicity of N2 is two & O3 is three.

(2)To find the relationship between molecular mass & vapour density of gas-(relative density)

Vapour density of gas=Density of gas/density of hydrogen

           =Mass of [certain vol of gas/same volume of H2] at STP

            =Mass of [n molecule of gas/ n molecule of H2] at STP

            =Mass of [1 molecule of gas/ 1 molecule of H2] at STP

 

                          Vapour density=Molecular Mass/2

(3)To find the relationship between mass & volume of gas-As the

Molecular Mass=Vapour density x 2

Or Molecular Mass=Mass of 22.4 L of gas at STP

Thus 22.4 L of any gas at STP weight is equal to the molecular mass of the gas expressed in grams which is called Gram-Molecular Volume Law (G.M.V.).

   Related Questions in Chemistry

  • Q : Lowering of vapour pressure Help me to

    Help me to go through this problem. Lowering of vapour pressure is highest for: (a) urea (b) 0.1 M glucose (c) 0.1M MgSo4 (d) 0.1M BaCl2

  • Q : Analytical chemistry 37% weight of HCl

    37% weight of HCl and density is 1.1g/ml. find molarity of HCl

  • Q : Question related to molarity Help me to

    Help me to go through this problem. Molarity of a solution containing 1g NaOH in 250ml of solution: (a) 0.1M (b) 1M (c) 0.01M (d) 0.001M

  • Q : Excel assignment I want it before 8 am

    I want it before 8 am tomorow please. I am just wondering how much is going to be ?

  • Q : Molecular basis of third law. The

    The molecular, or statistical, basis of the third law can be seen by investigating S = k in W.The molecular deductions of the preceding sections have led to the same conclusions as that stated in the third law of thermodynamics, namely, that a value can be

  • Q : Macromolecules what are condensation

    what are condensation polymerization give in with 2 examples

  • Q : Problem related to molarity Provide

    Provide solution of this question. Increasing the temperature of an aqueous solution will cause: (a) Decrease in molality (b) Decrease in molarity (c) Decrease in mole fraction (d) Decrease in % w/w

  • Q : Linde liquefaction process Liquefied

    Liquefied natural gas (LNG) is produced using a Linde liquefaction process from pure methane gas at 3 bar and 280 K (conditions at point 1 in figure below). A three-stage compressor with interceding is used to compress the methane to 100 bar (point 2). The first stage

  • Q : BASIC CHARACTER OF AMINES IN GAS PHASE,

    IN GAS PHASE, BASICITIES OF THE AMINES IS JUST OPPOSITE TO BASICITY OF AMINES IN AQEUOUS PHASE .. EXPLAIN

  • Q : Calculate molarity of a solution

    Provide solution of this question. Molarity of a solution prepared by dissolving 75.5 g of pure KOH in 540 ml solution is: (a) 3.05 M (b) 1.35 M (c) 2.50 M (d) 4.50 M